Digitized  by  the  Internet  Archive 

in  2007  with  funding  from 

Microsoft  Corporation 


http://www.archive.org/details/elementsofgeometOOwentrich 


ELEMENTS 


v  r       ,-  ./' " 

G  E  O  M  E  T  R  Y: 


<;.    \.  v  ENTWORTH,  A.  If., 

raomsoK  UK  MA  I  It 


^   OF  THE 


BOSTON : 

PUBLISHED   BY    GINN    AND    HEATH. 

L881. 


Copyright,  1877. 
By    GINN    AND    HEATH. 


I'kks.s  of  Rockwell  and  Churchill, 
39  Arch  St.,  Boston. 


PREFACE 


Most  persons  do  not  possess,  and  do  not  easily  acquire,  the 
powei  of  abstraction  requisite  for  apprehending  the  Geometri- 
cal conceptions,  and  for  keeping  in  mind  the  successive  steps 
of  a  continuous  argument     Bence,  with  a  very  large  proportion 

of  beginners  in  (leometry^it  depends  mainly  upon  the  form  in 
which  the  subject  i-  ]  vrhetner  they  pursue  the  study 

with  indifferen  o  say  aversion,  or  with  Increasing  i nt 

and  pleaeura 

In  oompiling  the  preeenl  treatise*  this  lad  has  1 n  kept  con- 
stantly in  view.  All  unnecessary  discussions  and  Bcholia  have 
been  avoided                h  methods  hare  been  adopted  aa  experi- 

comhined  with  repeated  trials, 
have  shown  to  be  most  readily  comprehended.  No  attempt  has 
been  made  to  render  more  intelligible  the  simple  notions  of 
position,  magnitude,  and  direction,  which  every  child  derives 
lVoin   obsi  :    hut   it   is  believed   that    these  notions  have 

nd  defined  with  mathematical  precision. 

A  few  symbols,  which  stand  tor  words  and  not  for  operations, 
have  been  used,  hut  these  are  of  so  great  utility  in  giving  style 
and  v  to  the   demonstrations  that  no  apology    seems 

for  their  introduction. 

(Ircat  pains  have  been  taken  to  make  the  page  attractive. 
The  Inures  are  large  and  distinct,  and  are  placed  in  the  middle 
of  the  page,  so  that  they  fall  directly  under  the  eye  in  imme- 
diate connection  with  the  corresponding  text.     The  given  lines 


PKEFACE. 


of  the  figures  are  full  lines,  the  lines  employed  as  aids  in  the 
demonstrations  are  short-dotted,  and  the  resulting  lines  are  long- 
dotted. 

In  each  proposition  a  concise  statement  of  what  is  given  is 
printed  in  one  kind  of  type,  of  what  is  required  in  another,  and 
the  demonstration  in  still  another.  The  reason  for  each  step 
is  indicated  in  small  type  between  that  step  and  the  one  follow- 
ing, thus  preventing  the  necessity  of  interrupting  the  process  of 
the  argument  by  referring  to  a  previous  section.  The  number 
of  the  section,  however,  on  which  the  reason  depends  is  placed 
at  the  side  of  the  page.  The  constituent  parts  of  the  propo- 
sitions are  carefully  marked.  Moreover,  each  distinct  assertion  in 
the  demonstrations,  and  each  particular  direction  in  the  construc- 
tions of  the  figures,  begins  a  new  line ;  and  in  no  case  is  it  neces- 
sary to  turn  the  page  in  reading  a  demonstration. 

This  arrangement  presents  obvious  advantages.  The  pupil 
perceives  at  once  what  is  given  and  what  is  required,  readily 
refers  to  the  figure  at  every  step,  becomes  perfectly  familiar  with 
the  language  of  Geometry,  acquires  facility  in  simple  and  accu- 
rate expression,  rapidly  learns  to  reason,  and  lays  a  foundation 
for  the  complete  establishing  of  the  science. 

A  few  propositions  have  been  given  that  might  properly  bo 
considered  as  corollaries.  The  reason  for  this  is  the  great  diffi- 
culty of  convincing  the  average  student  that  any  importance 
should  be  attached  to  a  corollary.  Original  exercises,  however, 
have  been  given,  not  too  numerous  or  too  difficult  to  discourage 
the  beginner,  but  well  adapted  to  afford  an  effectual  test  of  the 
degree  in  which  he  is  mastering  the  subjects  of  his  reading. 
Some  of  these  exercises  have  been  placed  in  the  early  part  of 
the  work  in  order  that  the  student  may  discover,  at  the  outset, 
that  to  commit  to  memory  a  number  of  theorems  and  to  repro- 
duce them  in  an  examination  is  a  useless  and  pernicious  labor; 
but  to  learn  their  uses  and  applications,  and  to  acquire  a  readi- 
ness in  exemplifying  their  utility,  is  to  derive  the  full  benefit 
of  that  mathematical  training  which  looks  not  so  much  to  the 


1' KM  FACE. 


M  to  the  'dudplvu  of  the  mental  fac- 

It  only  remains  to  express  my  sense  of  obligation  to  Dr. 

I).    F.    \Vi:u>    fox   valuable   assistance,    and    to  the   University 

for  the  irith  which  the  book  has  been  printed; 

and  also  to  give  assurance   tliat  any  suggestions  relating  to  the 

work  will  be  thankfully  received. 

G.  A.  WENTWOKTH. 
Phillips  Kxkter  Acadi 
January,  1878. 


NOTE  TO  THIRD   EDITION 

In  this  edition   I   have  ei  i  more  rF 

but   not  less  simple,  treatment  dlela,    Ratio,   and 

Limits.     The  ehangea  are  not  sufficient  to  prevent  the  Bimultar 

ia  use  of  the  old  and  new  editiona  in  the  data  :  still  the] 
very  important,  and  have  been  made  after  the  most  careful  and 
prolonged  conai 

I  have  to  express  my  thanks  for  valuable  suggestions  received 
in  my  correspondents;  and  a  special  nt  is  due 

from  me  to  Professor  C.   EL  Judson,  of  Furman   [Jnivei 
Greenville,  South  Carolina,  to  whom  I  am  indebted  for  assist- 
aiice  in  effecting  many  improvemente  in  thia  edition. 

TO  THE  TEACHER. 

When  the  pupil  Lb  reading  eacli  Book  for  the  Brat  time,  it  will  be 
well  to  let  him  write  hia  proofa  <>n  the  blackboard  in  hia  own  Ian- 
re  being  taken  that  hia  language^  be  the  sunniest  possible, 

that  the  arrangement  of  work  be  vertical  (without  side  work),  and 
that  the  figures  be  accurately  constructed. 

This  method  will  furnish  a  valuable  exercise  as  a  language  lesson, 
will  cultivate  the  habit  of  neat  and  orderly  arrangement  of  work, 
and  will  all«>w  a  brief  interval  for  deliberating  on  each  Btep, 

After  a  Book  has  been  read  in  thia  way  the  pupil  should  review 
Book,  and  should  be  required  to  draw  the  figures  free-hand.    He 


PKEFACE. 


should  state  and  prove  the  propositions  orally,  using  a  pointer  to 
indicate  on  the  figure  every  line  and  angle  named.  He  should  be 
encouraged,  in  reviewing  each  Book,  to  do  the  original  exercises  ;  to 
state  the  converse  of  propositions  ;  to  determine  from  the  statement, 
if  possible,  whether  the  converse  be  true  or  false,  and  if  the  converse 
be  true  to  demonstrate  it ;  and  also  to  give  well-considered  answers 
to  question's  which  may  be  asked  him  on  many  propositions. 

The  Teacher  is  strongly  advised  to  illustrate,  geometrically  and 
arithmetically,  the  principles  of  limits.  Thus  a  rectangle  with  a 
constant  base  b,  and  a  variable  altitude  x,  will  afford  an  obvious 
illustration  of  the  axiomatic  truth  contained  in  [4],  page  88.  If  x 
increase  and  approach  the  altitude  a  as  a  limit,  the  area  of  the  rec- 
tangle increases  and  approaches  the  area  of  tKe  rectangle  a  b  as  a 
limit ;  if,  however,  x  decrease  and  approach  zero  as  a  limit,  the  area 
of  the  rectangle  decreases  and  approaches  zero  for  a  limit.  An  arith- 
metical illustration  of  this  truth  would  be  given  by  multiplying  a 
constant  into  the  approximate  values  of  any  repetend.  If,  for  exam- 
ple, we  take  the  constant  60  and  the  repetend  .3333,  etc.,  the  approxi- 
mate values  of  the  repetend  will  be  T8^-,  fifc,  i%s$j,  ^Atf*  etc->  ancl 
these  values  multiplied  by  60  give  the  series  18,  19.8,  19.98,  19.998, 
etc.,  which  evidently  approach  20  as  a  limit ;  but  the  product  of  60 
into  J  (the  limit  of  the  repetend  .333,  etc.)  is  also  20. 

Again,  if  Ave  multiply  60  into  the  different  values  of  the  decreasing 
series,  ^,  ^-,  ^^  sooocp  etc->  which  approaches  zero  as  a  limit, 
we  shall  get  the  decreasing  series,  2,  £,  -gV>  jhfr  etc-  >  ail(l  tn*s  series 
evidently  approaches  zero  as  a  limit. 

In  this  way  the  pupil  may  easily  be  led  to  a  complete  comprehen- 
sion of  the  whole  subject  of  limits. 

The  Teacher  is  likewise  advised  to  give  frequent  written  examina- 
tions. These  should  not  be  too  difficult,  and  sufficient  time  should 
be  allowed  for  accurately  constructing  the  figures,  for  choosing  the 
best  language,  and  for  determining  the  best  arrangement. 

The  time  necessary  for  the  reading  of  examination-books  will  be 
diminished  by  more  than  one-half,  if  the  use  of  the  symbols  employed 
in  this  book  be  permitted. 

G.  A.  W. 

Phillips  Exeter  Academy, 
January,  1879. 


CONTENTS. 


PLANE   GEOMETRY. 

BOOK  I.    Eta -tii.ini-.ai:  l  Paob 

bmtOPUOTOmT  Remarks 3 

00110*1 4 

Strak.iii    Links         .         .     • G 

J'i. am.   An«;i.ks 7 

Angular  M acmitde D 

IMPOSITION 10 

Math  km  a  i  km.  Tnjn 11 

Axioms  and  Postulates IS 

Symbols  a  1  :j 

i«  -ULAR    AM)    OM  IS 14 

Pabaiul  lam 24 

Triangles 37 

i.s 58 

POLYQOOT  al 68 

BOOK   II.    Circles. 

Dkkinhk»ns 73 

SiKAK.nr  Lam  and  Circles 75 

Measurement 86 

Theory  of  Limits 87 

SUPPLKMI  MARY    PROPOSITIONS 100 

i  ructions 103 

BOOK  III.    Proportional  Lines  and  Similar  Polygons. 

Theory  of  Proportion 128 

Proportional  Lines 139 

Similar  Polygons 143 

era    .        .        .     • 164 


Mil  CONTENTS. 


BOOK  IV.    Comparison  and  Measurement  of  the  Sur- 
faces of  Polygons. 

Comparison  and  Measurement  of  Polygons       .        .        .174 

Constructions 194 

BOOK  V.    Begular  Polygons  and  Circles. 

Regular  Polygons  and  Circles 210 

Constructions 224 

ISOPERIMETRICAL   POLYGONS.       SUPPLEMENTARY          .            .            .  237 

Symmetry.     Supplementary 245 


ELEMENTS  OF  GEOMETRY. 


OP  THE 

JWIVERSITT] 

BOOK    I. 

RECTILINEAR  FIGURES. 


Introductory  1 1  km  arks. 

A  BOUOB  Mock  of  marble,  under  the  stone-cutter's  hammer, 
may  1  I  assume  regularity  of  form. 

H  a  block  be  eat  in  the  shape  repre-  , / 

I  in  this  diagram, 

It  will  have  six  flat  faces. 

Each  face  of  the  block  is  called  a  Sur-  /' 

face.  I  *' 

It  these  surfaces  be  made  smooth  by  pol- 
ishing, m  that,  vrhen  a  sti  ge  is  applied  t<>  any  cue  of 
them,  the  straight-edge  in  every  pari  will  touch  the  surface,  the 
«  s  are  called  Plane  Surfaces. 

The  sharp  edge  in  which  any  two  of  these  surface 
called  a  I 

Ihe  place  aft  which  any  three  of  these  lines  meet  is  called  a 
. 

I  t  now  the  block  be  removed,  we  may  think  of  the  place 

pied  by  the  block  as  being  of  precisely  the  same  shape  and 
lock  itself;  also,  as  having  surfaces  or  boundaries 
whicl  bom  surrounding  space.     AVe  may  likewise 

think  of  these  surfaces  as  having  lines  for  their  boundaries  or 
limits  ;  and  of  these  lines  as  having  points  for  their  extremities 
or  limits. 

A  Solid,  as  the  term  is  used  in  Geometry,  is  a  limited  por- 
tion of  space. 

r  we  acquire  a  clear  notion  of  surfaces  as  boundaries  of 
solids,  we  can  easily  conceive  of  surfaces  apart  from  solids,  and 


GEOMETRY. BOOK    I. 


suppose  them  of  unlimited  extent.  Likewise  we  can  conceive  of 
lines  apart  from  surfaces,  and  suppose  them  of  unlimited  length; 
of,  points  apart  from  lines  as  having  position,  but  no  extent 

Definitions. 

1.  Def.  Space  or  Extension  has  three  Dimensions,  called 
Length,  Breadth,  and  Thickness. 

2.  Def.    A  Point  has  position  without  extension. 

3.  Def.  A  Line  has  only  one  of  the  dimensions  of  exten- 
sion, namely,  length. 

The  lines  which  we  draw  are  only  imperfect  representations 
of  the  true  lines  of  Geometry. 

A  line  may  be  conceived  as  traced  or  generated  by  a  point  in 
motion. 

4.  Def.  A  Surface  has  only  two  of  the  dimensions  of  ex- 
tension, length  and  breadth. 

A  surface  may  be  conceived  as  generated  by  a  line  in  motion. 

5.  Def.  A  Solid  has  the  three  dimensions  of  extension, 
length,  breadth,  and  thickness.  Hence  a  solid  extends  in  all  direc- 
tions. 

A  solid  may  be  conceived  as  generated  by  a  surface  in  motion. 

Thus,  in  the  diagram,  let  the  upright 
surface  A  B  CD  move  to  the  right  to 
the  position  E  F  II  K.  The  points 
A,  B,  C,  and  D  will  generate  the  lines 
AE,  BF,  CK,  and  D II  respectively.         C  K 

And  the  lines  A  B,  B D,  DC,  and  A  C  will  generate  the  sur- 
faces A  F,  B II,  D  K,  and  A  K  respectively.  And  the  surface 
ABC D  will  generate  the  solid  A  H. 

The  relative  situation  of  the  two  points  A  and  H  involves 
three,  and  only  three,  independent  elements.  To  pass  from  A  t<  I  // 
it  is  necessary  to  move  East  (if  we  suppose  the  direction  A  E  to 


I 

I 

J// 


DEFINITIONS. 


be   i  i  distance  equal  to  A  E,  North  a  distance  equal  to 

/•//',  and  down  a  distance  equal  to  F II. 

These  three  dimensions  we  designate  for  convenience  length, 

Lth,  and  thickness. 

6.  Tin-  limits  (extremitiee)  of  lines  are  points. 
The  limits  (houndariee)  of  surfaces  an  tinea* 
The  limits  (boundaries)  of  solids  are  surfaces. 

7.  I  >i:i  .     Extension  is  also  called  Magmtt 

When  reference  is  had  to  extent,  lines,  .  and  solids  are 

called  ma 

8.  Def.    A  Straight  line  is  a  line  which  has 
the  same  direction  throughout  its  who) 

!».   I>rr.  X  Curved  line  which  changes 

its  directi  int. 

10.  I'm      A  Broken  line  is  a  series  of  con- 

i:t  lines. 
When  the  word  line  is  used  a  straight  lino  is  meant,  and 
when  the  woid  eui  I  line  is  meant. 

11.  1  >i  i  .     A  Flam  St  in  which, 
if  any  two  points  be  taken,  the  straight  line  joining  these  points 

will  lie  wholly  iii  the  surface. 

1 -.    l»i:i.    A  «*  is  a  surface  no  part  of  which 

is  plana 

13.  Figure  or  form  depends  upon  the  relative  position  of 
points.     Thus,  the  figure  oi  form  of  a  line  (straight  or  curved) 

depends   upon   the   relative   position   of  points   in  that  line;   the 

figure  or  form  of  a  Burface  depends  upon  the  relative  position  of 
points  in  that  surface. 

When  reference  is  had  to  form  or  shape,  lines,  surfaces,  and 

solids   are   .-all    | 


G  GEOMETRY. BOOK   I. 

14.  Def.  A  Plane  Figure  is  a  figure,  all  points  of  which 
are  in  the  same  plane. 

15.  Def.  Geometry  is  the  science  which  treats  of  position, 
magnitude,  and  form. 

Points,  lines,  surfaces,  and  solids,  with  their  relations,  are 
the  geometrical  conceptions,  and  constitute  the  subject-matter  of 
Geometry. 

16.  Plane  Geometry  treats  of  plane  figures. 

Plane  figures  are  either  rectilinear,  curvilinear,  or  mixtilinear. 

Plane  figures  formed  by  straight  lines  are  called  rectilinear 
figures ;  those  formed  by  curved  lines  are  called  curvilinear  fig- 
ures ;  and  those  formed  by  straight  and  curved  lines  are  called 
mixtilinear  figures. 

17.  Def.  Figures  which  have  the  same  form  are  called 
Similar  Figures.  Figures  which  have  the  same  extent  are  called 
Equivalent   Figures.     Figures  which  have  the  same  form  ami 

1  are  called  Equal  Figures. 


On  Straight  Lines. 

18.  If  the  direction  of  a  straight  line  and  a  point  in  the 
line  be  known,  the  position  of  the  line  is  known;  that  is,  a 
straight  line  is  determined  in  position  if  its  direction  and  one  of 
its  points  be  known. 

Hence,  all  straight  lines  which  pass  through  the  same  point  in 
the  same  direction  coincide. 

Between  two  points  one,  and  but  one,  straight  line  can  be 
drawn ;  that  is,  a  straight  line  is  determined  in  position  if  turn  of 
its  points  be  known. 

Of  all  lines  between  two  points,  the  shortest  is  the  straight 
line;  and  the  straight  line  is  called  the  distance  between  the 
two  points. 


]»K1  INITIONS. 


The  point  from  which  a  line  isjdrawn  is  called  its  origin. 

19.  II'  B  line,  as  C  B,  A  f  Bj  be  produced  through  C, 
the  portions  C  B  and  C  A  may  be  regarded  as  different  lines 
having  opposite  directions  from  the  point  C\ 

Hence,  even  line,  as  A  B,  t f,  has  two  opposite 

directions,  namely  from  A  toward  B,  which  is  expressed  by  say- 
ing  line  A  B,   and    from   B  toward  A,   which  is  expressed  by 
line  A'  A. 

20.  If  a  straight  line  change  its  magnitude,  it  must  become 

!•  or  shorter.     Thus  by  prolonging  A  B  to  C,  - f ?, 

-  A  11  +  B  C ;  and  conversely,  /;('=.!('-;(  B. 

If  a  line  iia-rease  so  tliat  it  is  prolonged  by  its  own  magnitude 
several  times  in  succession,  the  line  is  muk  suit- 

ing line  is  called  a                   oi  the  given  line.     Thus,  if  A  B  = 
B(   -CD,   etc.,   A    f      \ £_£,    then   AC      'I  A  /;,    AD  = 

:\A  /;. 

It  must  also  be  possible  to  divide  B  il  line  into  an 

ued  number  of  equal  parte.     For,  assumed  thai  the  *th 

part  of  a  given  line  were  not  attainable,  then  the  double,  triple, 

quadruple,  of  the  *th  part  would  imt  be  attainable.     Among 

these  multiples,  however,  we  should  reach  the  nth  multiple  of 

this  ath  part,  that  is,  the  line  itself      Hence,  the  line  itself  would 

not  be  attainable  ;  which  contradicts  the  hypothesis  that  we  have 
ven  line  before  us. 
Therefore,  U  i.<<ihr,t,t  to  add,  subtract,  multiply,  and 

divide  Kne$  of  ith. 

21.  Sin  traight  line  has  the  property  of  direction, 
it   must    he    true    that   two  straight  lines  have  either  the  same 

or  <////'  /•'  ns. 

Two  ttraigkt  Une$  wkich  have  the  same  direction,  without  coin- 
cide* nit ;  for  if  they  could  meet,  then  we  should 
have  two  straight  lines  passing  through  the  same  point  in  the 
same  direction.      Such  lines,  however,  coincide.  §  18 


8  GEOMETRY. BOOK    I. 

22.  Two  straight  lines  which  lie  in  the  same  plane  and  haw 
(liferent  directions  must  meet  if  sufficiently  prolonged  ;  and  micst 
have  one,  and  but  one,  point  in  common. 

Conversely  :  Two  straight  lines  lying  in  the  same  plane  which 
do  not  meet  have  the  same  direction;  for  if  they  had  different 
directions  they  would  meet,  which  is  contrary  to  the  hypothesis 
that  they  do  not  meet. 

Two  straight  lines  which  meet  have  different  directions;  for 
if  they  had  the  same  direction  they  would  never  meet  (§  21), 
which  is  contrary  to  the  hypothesis  that  they  do  meet. 


Ox  Plane  Angles. 

23.  Def.  An  Angle  is  the  difference  in  direction  of  two 
lines.  The  point  in  which  the  lines  (prolonged  if  necessary) 
meet  is  called  the  Vertex,  and  the  lines  are  called  the  Sides  of 
the  angle. 

An  angle  is  designated  by  placing  a  letter  at  its  vertex,  and 
one  at  eacli  of  its  sides.      In  reading,  we  name  the  three  let- 
ters, putting  the  letter  at  the  vertex  between  the  other  two.    Win  mi 
the  point  is  the  vertex  of  but  one  angle  we  usually  name  the. 
r  at  the  vertex  only;  thus,  in  Fig.  1,  we  read  the  angle  by 


D 


H 

Fig.  1.  Fig.  2. 

calling  it  angle  A.  But  in  Fig.  2,  If  is  the  common* vertex  of 
two  angles,  so  that  if  we  were  to  say  the  angle  //,  it  would  not 
be  known  whether  we  meant  the  angle  marked  3  or  that 
marked  4.  We  avoid  all  ambiguity  by  reading  the  former  as 
tie*  angle  K  II  />,  and  the  latter  as  the  angle  E H F. 


hKFIXITIONS. 


D 


The  magnitude  of  an  angle  depends  wholly  upon  the  extent 

i   of  its   sides,  and  not  upon  their 

length.     Thus  if  the  sides  of  the  angle  BAG, 

namely,   A  B   and   A  G,  be  prolonged,  their 

extent  of  opening  will  not  be  altered,  and  the 

of  the  angle,  consequently,  will  not  be 
chan 

24.  Dkf.    Adjacent  Anglet  are  an 

bavin,  mon  vertex  and  a  common 

side   between    them.      Tims    the  angles 
G  D  E  and  G  D  F  are  adjacent  angles.         E 

25.  Def.  A  Right  Angle  is  an  angle  included  between  two 

lit    lines    which   meet   each    other  SO   that    the   two  adjacent 

angles  Formed  by  producing  one  of  the  tin 

through  '  Thus    it'  the 

straight  line  A  B  meal  theatraighl  line  G D 
so  tint  ;1  ut  angles  ABC  and  ABB 

are  equal  to  one  another,  each  of  these  an- 
gles is  call  t  angle. 

26.  DlP.  Perpen 
which  make  a  righl  angle  with  each  other. 

27.  Def.   An  A<  is  an  an 

BA  G. 
ie  is  an  an 
iter   than   a   right   angle;    as  the  angle 

p  /;  r. 

29.  ngles,  in 

distinction  from  right  angles,  are  called  ob- 


B 


'D 


C 


-D 


rlet;  and  intersecting. lines  which  are  not  perpendicular 
to  each  other  are  called  line*,  A 

30.  Def.    The  Complejnent  of  an  angle  is 
the  difference  between  a  right  angle  and  the 
n  angle.     Tims  A  B  D  is  the  complement 
of  the  angle   D  BO\  also  I)  BC  is  the  com- 
plement pf  the  angle  A  B D. 


10 


GEOMETRY.  —  BOOK   I. 


31.  Def.  The  Supplement  of  an  angle 
is  the  difference  between  two  right  angles 
and  the  given  angle.  Thus  A  CD  is  the 
supplement  of  the  angle  D  C  B;  also  D  C  B 
is  the  supplement  of  the  angle  AC  D. 

32.  Def.  Vertical  Angles  are  angles 
which  have  the  same  vertex,  and  their 
sides  extending  in  opposite  directions. 
Thus  the  angles  AOD  and  COB  are 
vertical  angles,  as  also  the  angles  A  0  C 
and  DOB. 


D 


-D 


^  On  Angular  Magnitude. 


A'- 


B' 


A 


33.  Let  the  lines  B  B'  and  A  A'  be  in  B 

the  same  plane,  and  let  B  B'  be  perpen- 
dicular to  A  A'  at  the  point  0. 

Suppose  the  straight  line  0  C  to  move 
in  this  plane  from  coincidence  with  0  A, 
about  the  point  0  as  a  pivot,  to  the  po- 
sition 0  C ;  then  the  line  0  C  describes  or 
generates  the  angle  A  0  C. 

The  amount  of  rotation  of  the  line,  from  the  position  0  A  to 
the  position  0  C,  is  the  Angular  Magnitude  A  0  C. 

If  the  rotating  line  move  from  the  position  0  A  to  the  po- 
sition 0  B,  perpendicular  to  0  A,  it  generates  a  right  angle  ;  to 
the  position  0  A1  it  generates  two  right  angles ;  to  the  position 
0  Bf,  as  indicated  by  the  dotted  line,  it  generates  three  right 
angles ;  and  if  it  continue  its  rotation  to  the  position  0  A, 
whence  it  started,  it  generates  four  right  angles. 

Hence  the  whole  angular  magnitude  about  a  point  in  a  plane 
is  equal  to  four  right  angles,  and  the  angular  magnitude  about 
a  point  on  one  side  of  a  straight  line  drawn  through  that  point 
is  equal  to  two  right  angles. 


DEFINITIONS. 

11 

c 

^ 

\ 

/ 

/B 

\ 

D 

7 

K 

A 

A 

A 

£» 

/ 

If 

N> 

/> 

11 

0 

A 

Fig 

1. 

2. 

:;i.  ice  the  an.  pritude  about  the  point  0  is 

bet  increased  nor  diminished  by  the  number  of  lines  which 
radiate  from  thai   poinl .  i  of  <>!l  tl 

in  a  plane,  as  A  0  B  +  B  0  (7  +  C  0  D,  etc,  in  Fig.  1,  is  equal 

;  an«  I  the  sum  of  all  U 
on   one  side  of  a  straight    lint  dra  ih   that 

A  <>  /;  +  /;  o  C  +  CO  I>.  '  to  two  right 

Henoe  twn  tdjacenJ  angles,  OCA  and  OCB,  jj 

formed  by  two  itraighl   linesj  of  which  one  is 
produced  from  the  pew  in  both  di- 

rects: 

l)e  called 


Ox  mi    Method  of  Superposition'. 

of  the  equality  of  two  geometrical  magnitudes 
IS  that  they  coincide  point  for  point, 

Thus,  two  equal,  if  they  can  be  so  placed 

thai  tli*-  points  at   fcheii  tities  coincide.     Two  angles  are 

equal,  if  they  can  laced  that  their  vertices  coincide  in 

tion  and  their  sides  in  direction. 

In  applying  this  test  of  equality,  we  assume  that  a  line  may 

be  moved  from  one  place  t<>  another  without  altering  its  length; 

may   he  taken   up,    turned  over,  and  put  down, 

without  altering  the  difference  in  direction  of  its  sides. 


12 


GEOMETRY. BOOK    I. 


D 


This  method  enables  us  to  com- 
pare unequal  magnitudes  of  the 
same  kind.  Suppose  we  have  two 
angles,  ABC  and  A'  B'  C  Let 
the  side  B  C  be  placed  on  the  side 
B'  C,  so  that  the  vertex  B  shall  fall  on  B',  then  if  the  side  B  A 
fall  on  BlAr,  the  angle  ABC  equals  the  angle  A'  B'  C ;  if  the 
side  B  A  fall  between  B'  C  and  B'  A'  in  the  direction  B'  D,  the 
angle  ABC  is  less  than  A'  B'  6"  ;  but  if  the  side  B  A  fall  in  the 
direction  B'E,  the  angle  A  B  C  is  greater  than  A'  Bf  C. 

This  method  of  superposition  en-  g  C 

ables  us  to  add  magnitudes  of  the 
same  kind.     Thus,  if  we  have  two     c 
straight   lines    AB  and    CD,    by     A 
placing  the  point  C  on  B,  and  keeping  C  D  in  the  same  direc- 
tion with  A  By  we  shall  have  one  continuous  straight  line  A  D 
equal  to  the  sum  of  the  lines  A  B 
and  C  D. 

in  :  if  we  have  the  angles 
A  11  0  and  D  E  F,  by  placing 
the  vertex  B  <>n  K  and  the  side 
BC  in  the  direction  of  ED,  the 
angle  AB  C  will  take  the  position 
A  ED,  and  the  angles  D  E F  and 
J  B  C  will  together  equal  the  an- 
gle A  EF. 


-D 
-B 


Mathematical  Terms. 

36.  Def.   A  Demonstration  is  a  course  of  reasoning  by  which 
the  truth  or  falsity  of  a  particular  statement  is  logically  established. 

37.  Def.    A  Theorem  is  a  truth  to  be  demonstrated. 

38.  Def.    A  Construction  is   a  graphical  representation  of 
a  geometrical  conception. 

39.  Def.    A  Problem  is  a  construction  to  be  effected,  or  a 
question  to  be  investigated. 


DEFINITIONS.  13 


40.  Def.  An  Axiom  is  a  truth  which  is  admitted  without 
demonstration. 

41.  DsF.  A  Postulate  is  a  problem  which  is  admitted  to 
I*.-  possible. 

42.  DSP.    A  /V<  7  is  i  it  her  a  theorem  or  a  problem. 

43.  Ih.i.  A  Corollary  is  a  truth  easily  deduced  from  the 
proposition  to  which  it  is  attached. 

1  1.  Def.  A  Scholium  is  b  remark  upon  some  particular  fea- 
ture of  a  proposition. 

45.  Ihr.  An  Hypothesis  is  a  supposition  made  in  the 
enunciation  <>f  a  proposition,  ox  in  the  course  of  a  demonstration. 

Axiom. 

1.   Things  which  are  equal  to  the  same  tiling  arc  equal  to  each 

oti 

•_'.  When  equals  arc  added  to  equals  the  sums  are  equal. 

3.  When  equals  arc  taken  I  are  equal. 

1.  When  equals  are  added  to  iiiicquals  I  arc  unequal 

5.  When  equals  are  taken  bom  anequala  the  remainders 

■iual. 

G.  Things  which  arc  double  tl  thing,  or  equal  things, 

re  equal  to  each  oth< 

7.  Thin   a    which   are  halves   of  the  same  thing,   or  of  equal 

tliih  ch  other. 

8.  The  w  r  than  any  of  its  parts. 

'.».   The   wlml  .1  to  all  its  parts  taken  together. 

17. 

ated  — 

1.  That  a  straight  line  can  be  drawn  from  any  one  point  to  any- 
other  point. 

l\  Thai  a  straight  line  can  be  produced  to  any  distance,  or  can 
be  terminated  at  any  point. 

3.  That  the  circumference  of  a  circle  can  be  described  about  any 
i  any  distance  from  that  centre. 


14 


GEOMETRY. BOOK   I. 


48.  Symbols  and  Abbreviations. 


.*.  therefore. 
=  is  (or  are)  equal  to. 
Z  angle. 
A  angles. 
A  triangle. 
A  triangles. 
II   parallel, 
O  parallelogram 
HJ  parallelograms. 
_L  perpendicular. 

Jl  perpendiculars. 

rt.  Z  right  an 

it  angles. 

>  is  (or  are)  greater  than. 

<  is  (or  are)  less  than. 

rt.  A  right  triangle. 

rt.  A  right  triangli 

O  circle. 

©  circles. 

+  increased  by. 

—  diminished  by. 

X  multiplied  by. 

-4-  divided  by. 


Post,  postulate. 
Def.  detinition. 
Ax.  axiom. 
Hyp.  hypothesis. 
Cor.  corollary. 

Q.  E.  D.  quod  erat  demonstran- 
dum. 
Q.  E.  F.  quod  erat  faciendum. 
Adj.   adjacent. 
K\t.-int,   exterior-interior. 
Alt. -int.  alternate-interior. 
[den.  identical. 
( 'mis.   construct  ion. 
Sup.  supplementary. 
Sup.  adj.    supplcinentary-adja- 

Ex.  exercise. 
111.  illustration. 


PERPENDICULAR    AND    OBLIQUE    LINES.  15 

•x  Perpendicular  axd  #blique  Lixes. 

ri:tr#srn#x   I.     The#kkm. 

49.    When  one  ttoraigkt  line  crosses  another  straight  line 
He  *  til. 

o 


D 


I' 

Let  line  OP  cross  A  B  at  C. 

We  are  to  prove      Z  OCB  =  Z  A  CP. 

Z  0  C  A  +  Z  0  CB  =  2  rt.  A,  §34 

10  sup. -adj.  A). 

Z  OCA  +  Z  A  CP  =  2  rt.  A,  §  34 

i.A). 

.    .     <>< '  A  +  ZOC  B  =  ZOCA  +  Z  AC  P.     Ax.  1. 

Tab  of  these  equals  the  common  ZOC  A. 

Tn  ZOCB  =  ZACP. 

In  like  mam,  v  prove 

Z  ACO  =  Z  PCB. 

Q.  E.  D. 

50.  COROLLARY.    If  two  straight  lines  cut  one  another,  the 
four  angles  which  they  make  at  the  point  of  intersection  are 
fchei  equal  to  four  right  angles. 


16  GEOMETRY.  —  BOOK    I. 

Proposition  II.     Theorem. 

51.  When  the  sum  of  two  adjacent  angles  is  equal  to  two 
right  angles,  their  exterior  sides  form  one  and  the  same 
straight  line. 

0 


Let  the  adjacent  angles  Z  OCA  +  Z  0  C  B  =  2  it.  A. 
We  are  to  prove  A  C  and  C  B  in  the  same  straight  line. 
Suppose  C  F  to  be  in  the  same  straight  line  with  A  C. 

Thru  Z  OCA  +  Z  OCF=  2  it.  A.  §34 

[being  sap. -adj.  A ). 

But  Z  OCA  +  Z  OCB  =  2  rt.  A.  Hyp. 

.-.Z  OCA  +  Z  OC F=Z  OCA  +  Z  OCB.     Ax.  1. 

Take  away  from  each  of  these  equals  the  common  Z  0  C  A . 

Then  Z  OCF=Z  OCB. 

.'.  C  B  and  C F  coincide,  and  cannot  form  two  lines  as  rep- 
resented in  the  figure. 

.  * .  A  C  and  C  B  are  in  the  same  straight  line. 

Q.  E.  D. 


PBRPENDICULAB    AND    OBLIQUE    LINES. 


17 


Proposition  III.     Theorem. 
52.  A  perpentlh'iihir  measure*  the  shortest  distance  from 
a  /jo i uf  to  a  tiraigkt  1% 


i 
Let  A  B  be  the  given  straight  line,  C  the  given  point 
and  CO  the  perpendicular. 

\Y>  em  to  prove  C  O  <  drown  from  CtoA  />, 

as  C 

luce  CO  to  E,  making  OE=  CO. 

Drav,    A'/: 
On  A  B  U  a:  0  (."'  /'  until  it  comes  into  the 

plane  rfO  A'/'. 

The  line  0  C  will  take  the  direction  of  0  Ey 

■■<•  Z  COF=  ZEOF,  eachbeing  a  rt.  Z). 

The  point  0  will  fall  upon  the  poinl 

(*i?we  0  C  =  OE  by  cons.). 

.'.  line  CF"*lu*FB, 

(Afl  hi  the  same  points). 

(70+  0E=2  CO. 
C0+  OE<CF+  FE, 

(<>  d  is  the  shortest  distance  between  two  points). 

Substitute  2  C  0  for  CO  +  OF, 

an.l  2  CFfor  C  F+  F  E  ;  then  wo  have 

2  C0<  2(7>. 

Q    E.  D. 


and 
But 


§  18 

Cons. 
§  18 


18 


GEOMETRY. BOOK   I. 


Proposition  IV.     Theorem. 

53.  Two  oblique  lines  drawn  from  a  point  in  a  perpen- 
dicular, cutting  off  equal  distances  from  the  foot  of  the  per- 
pendicular, are  equal. 


Let  F  C  be  the  perpendicular,  and  C  A  and  C  0   two 
oblique  lines  cutting  off  equal  distances  from  F. 


We  are  to  prone         C  A  =  C  0. 

Fold  over  C  FA,  on  C F  as  an  axis,  until  it  comes  into  the 
plane  ofCFO. 

FA  will  take  the  direction  of  FO, 
(since  ZCF A  —  ZCFO,  each  being  art.  Z). 

Point  A  will  fall  upon  point  0, 

(FA  =  FO,  by  hyp.). 


.'.line  C A  =  line  CO, 
(their  extremities  being  the  same  points). 


§  18 


Q.  E    O. 


PKKI'KNDK  IIAR    AND    OBLIQUE    LINES.  19 


Proposition   V.     Theokkm. 

54.   The  sum  of  two  lines  drawn  from  a  point  to  the  ex- 
ities  of  a  straight  lii  iter  than  tie  sum  of  two 

oihrr  I  .  but  included  by  them. 


Let  C  A  and  Q  /J  be  two  lines  drawn  from  the  point  0 
to  the  extremities  of  the  straight  line  A  ft  Let  O  A 
and  0  B  be  two  lines  similarly  drawn,  but  included 
byCA  and  0  ft 

i  prove      CA  +  CB>OA  +  OB. 

Pru  1  E. 

Tl  AG+CB>AO+0£,  §18 

and  0  /■>  BO.  §  18 

Add  .  and  we  b 

l+CE+BE+0E>0A  +  0E+0B. 
Substitute  for  CE  +  BE  its  equal  C B, 
and  take  an  side  of  the  inequality. 

We  have         C-4  +  CB>0'A  +  OB. 

Q.  E.  D. 


20 


GEOMETRY.  - 


BOOK   I. 


Proposition  VI.     Theorem. 

55.   Of  two  oblique  lines  drawn  from  the  same  point  in  a 
perpendicular,  cutting  off  unequal  dista?ices  from  the  foot  of 
the  perpendicular,  the  more  remote  is  the  greater. 
C 


Let  C  F  be  perpendicular  to  A  B,  and  C  K  and  C  II  two 
oblique  lines  cutting  off  unequal  distances  from  F. 

1 1 V  are  to  prove  C II  >  C  K. 

Produce  CF  to  Ef  making  F  E=  C  F. 
Draw  JTJTand  K1L 
CII  =  //  h\  and  C  K  =  KE,  §  53 

(two  oblique  lines  drawn  from  the  same  point  in  a  J_,  cutting  off  equal  dis- 
tances from  the  foot  of  the  ±,  are  equal). 

But  CII+  II E  >CK+  K  E,  §  54 

(7Jhe  sum  of  two  oblique  lines  drawn  from  a  point  to  tlie  extremities  of  a, 
straight,  line  is  greater  than  the  sum  of  two  other  lines  similarly  drawn, 
but  included  by  them)  ; 

.-.  2  CII>  2  CK) 
.'.  CII>CK. 

Q.  E.  D. 

56.  Corollary.  Only  two  equal  straight  lines  can  be  drawn 
from  a  point  to  a  straight  line ;  and  of  two  unequal  lines,  the 
greater  cuts  off  the  greater  distance  from  the  foot  of  the  perpen- 
dicular. 


II ERF KXDICULAR   AND    OBLIQUE    LINES.  21 

Pro  position   VII.     Theorem. 

57.  Two  equal  oblique  lines,  drawn  from  tie  same  point 
in  a  perpendicular,  cut  off  equal  distances  from  the  foot  of 
fie  perpendicular. 


Let  C  F  be  the  perpendicular,  and  C  E  and  C  K  be  two 
equal  oblique  lines  drawn  from  the  point  C. 

Wi  m  to  i  rove  /'A'. 

Fold  over  C FA  on  0  F  as  an  axis,  until  it  comes  into  the 

plan.-  of  C  /-'A'. 

The  line  FB  will  take  the  direction  /'A', 

(Z  CFE=  Z  C  F  A.  i  art.  Z). 

Then  the  point  E  mud  fall  upon  the  point  K  \ 

otherwise  one  of  these  oblique  lines  must  be  more  remote  from 

the  _L, 

and  .*.  greater  than  the  other;   which  is  contrary  to  the 

hypothesis.  §  55 

.\  F  /;==  FK. 

Q.  E.  D. 


22 


GEOMETRY.  —  BOOK   I. 


Proposition  VIII.     Theorem. 

58.  If  at  the  middle  point  of  a  straight  line  a  per  pen- 
lot  be  erected, 

I.  Any  point  in  the  perpendicular  is  at  equal  dis fauces 
from  the  extremities  of  the  straight  line. 

II.  Any  point  without  the  perpendicular  is  at  unequal 
distance*  from  the  extremities  of  the  straight  line. 


Let  P 11  be  a  perpendicular  erected  at  the  middle  oi 
the  straight  line  A  B,  O  any  point  in  PR,  and  C  any 
point  without  7V.'. 


I.  Draw  OA  and  OB. 
We  are  to  prove  0  A  —  0  B. 
Since                            PA  =  PB, 

OA  =  OB,  §53 

•hliqnc  lines  drawn  from  the  same  point  in  a  _L,  cutting  off  equal  dis- 
tances from  the  foot  of  the  _L,  are  equal). 

II.  BmwCA  and  C  B. 

We  are  to  prove  C  A  and  C  B  unequal. 

One  of  these  lines,  as  CA,  will  intersect  fche  _L. 
From  D,  the  point  of  intersection,  draw  1)  11. 


I'KKPKNDh  ll.AU    AND    OBLIQUE    LINES.  23 

DB  =  DA,  §  53 

[twt  oh'i'i'  "  _L,  cutting  off  equal  dis- 

fout  of  tJie  J_,  are  equal). 

CB<  CD+  DB,  §  18 

(a  sf  is  tlie  shortest  distance  between  two  points). 

Substitute  f<»r  D B  its  equal  DA,  then 
CB  <  CD+  DA. 
But  CD  +  DA  =  CA,  Ax.  9. 

..CB<CA. 

Q.  E.  D. 

59.  The  Locus  of  a  point  is  a  1  r  curved,  con- 
taining all  the  joints  which  possess  a  common  prop 

Thus,  tin-  perpendiculai  i  I  the  middle  of  a  straight 

line  is  the  locus  <>f  all  points  equallj  from  the  extreini- 

ties  of  tha!  imr. 

60.  Scholium.  Si  fcermine  the  position  of 
a  straight  line,  two  points  equally  distant  from  the  extremities 
of  a  straighl    Line  det  rmine  the  perpendicular  at  the  middle 

1>- »int    of  that    line. 


Ex,  1.    If  an  an  hat  is  its  complement? 

L\    If  an  and*'  he  a  right  its  supplement  ? 

3.    If  an  angle  he  £  of  a  right  angle,  what  is  its  complement? 
1.    If  an  angle  be  £  of  a  right  anjgle,  what  is  its  supplement? 
5.    Show  that  tlie  hisectors  of  ,xwo  vertical  angles  form  one 
and   the   same   straight   line. 

w  that   the  two  straighl  lines  which  bisect  the  two 
d  angles  are  perpendicular  to  each  other. 


24 


GEOMETRY. 


-BOOK   I. 


Proposition  IX.     Theorem. 

61.  At  a  point  in  a  straight  line  only  one  perpendicular 
to  that  line  can  be  drawn ;  and  from  a  point  without  a 
straight  line  only  one  perpendicular  to  that  line  can  be  drawn. 


AE 


/>* 

Fig.  1. 


D 


Let  B  A  (fig.  1)  be  perpendicular  to  C  D  at  the  point  B. 

We  are  to  prove  B  A  the  only  perpendicular  to  C  D  at  the 
pot Hi  B. 

If  it  be  possible,  let  B  E  be  another  line  J_  to  C  D  at  B. 
Then  Z  EBD  is  art.  Z.  §  20 

Bat  Z  AB1)  is  a  rt.  Z.  §  26 

.\Z  EBD  =  Z  ABD.  Ax.  1. 

That  18,  a  part  is  equal  to  the  whole  ;  which  is  impossible. 
In  like  manner  it  may  be  shown  that  no  other  line  but  B  A 
is_L  to  CD  at  B. 

Let  AB  (fig.  2)  be  perpendicular  to  C  D  from  the  point  A. 
We  are  to  prcm  A  B  the  only  _L  to  C  D  from  the  point  A. 

If  it  be  possible,  let  A  E  be  another  line  drawn  from  A  _L 
to  CD. 

Conceive  Z  A  E  B  to  be  moved  to  the  right  until  the  ver- 
tex E  falls  on  B,  the  side  E B  continuing  in  the  line  CD. 

Then  the  line  E  A  will  take  the  position  B  F. 

Now  if  A  E  be  _1_  to  C  D,  B  F  is  _L  to  C  D,  and  there  will 
be  two  J*  to  C  D  at  the  point  B ;  which  is  impossible. 

In  like  manner,  it  may  be  shown  that  no  other  linn  hut 
A  B  is  JL  to  CD  from  A.  Q  E  D 

62.  Corollary.  Two  lines  in  the  s-dinc  plane  perpendicular 
to  the  same  straight  line  have  the  same  direction;  otherwise 
they  would  meet  (§  22^),  ami  we  should  have  two  perpendicular 
liius  drawn  from  their  point  of  meeting  to  the  same  line ;  which 
is  impossible. 


*v 


PARALLEL   LINES] 


UlTIVSKS^ 


I  >\  Parallel  Lines. 

Parallel  Lines  are  straight  lines  which  lie  in  the  same 

plane  and  have  the  Bame  direction,  or  opposite  directions. 

rile!  lines  lie  in  the  Bame  direction,  when  they  are  on 
iame  Bide  of  the  straight  line  joining  their  origins. 
Parallel  lines  lie  in  opposite  directions,  when  they  are  on 
oppo  of  the  straight  line  joining  their  origins. 

64,     Tu  "Ot  meet.  §  21 

-////'  plane  perpendicular  to  a  given 
line  I  I  art  therefore  parallel, 

\  a  given  point  only  om  J><  drawn  par- 

allel  to  a  g  §  18 


line  EF  cu1  two  other  straight  lines  AB 
and  ODj  it  makes  with  those  lines  eight  angles,  to  which  par- 
ticular nai:  iven. 

The  angles  I.  l.  6,  7  are  called  Interior  angles. 

Tli  2,  3,  5,  8  are  called  Exterior  angles. 

The  pairs  of  angles  1  and  7,  4  and  6  are  called  Alternate- 

The  pairs  of  angles  2  and  8,  3  and  5  are  called  Alternate- 

The  pairs  of  angles  1  and  5,  2  and  o,  4  and  8,  3  and  7  are 

•  all  r  angles. 


26 


GEOMETRY.  —  BOOK   I. 


Proposition  X.     Theorem. 

67.    If  a  straight  line  be  perpendicular  to  one  of  two 
parallel  lines,  it  is  perpendicular  to  the  other, 

H 


-B 


M 


-N 


K 

Let  A  B  and  E  F  be  two  parallel  lines,  and  let  UK  be 
perpendicular  to  A  B. 

\Y>  are  t<>J,rove      II K  J_  to  E F. 

Through  C  draw  MN  JL  to  UK. 

Then  MN  is  I!  to  A  B.  §  or, 

(Two  lines  in  the  same  plane  ±  to  a  given  line  are  parallel). 
But  EF  is  II  ioAB,  llvp. 

.'.  E F  coincides  with  M N.  §  66 

vugh  Hi<  aanu  point  only  one  line  can  he  drawn  II  to  a  given  line), 

.\EFis±to  //  K, 

that  ia  //  K  is  JL  to  E  F. 

Q.  E.  D. 


PARALLEL   LINES.  27 


Proposition  XL     Theorem. 

68.  If  two  parallel  ttraight  lines  be  cut  by  a  third 

straight  line  the  alternate-interior  angles  are  equal. 

A  B 


Let  EF  and  OH  be  two  parallel  straight  lines  cut  by 
the  line  BC. 

We  are  to  prove  Z  B  =  Z  C. 

Through  0,  the  middle  point  of  B  C,  draw  A  Z)Xto  G  II. 

Then  A  D  is  likewise  _L  to  E  §  67 

(a  ti  ±  to  one  of  two  lis  is  ±  to  the  other), 

that  is,  CD  and  B  A  are  both  ±  to  A  D. 

Apply  tigure  COD  to  figure  £04  so  that  02)  shall  fall 

on  0 

Thru  0  6' will  fall  on  OB, 

'' 0  D  =  Z.  B  0  A,  bei,  A); 

and  point  G  will  tall  upon  2?,  * 

>ce  0  C  =  0  B  by  construction). 

Then        _L  02>  will  coincide  with  _i_  AM.  §  01 

vm  _L  to  tlmt  line  can  be  drawn), 

.*.  Z  0  CD  coincides  with  Z  0  BA,  and  is  equal  to  it. 

Q.  E.  D. 

s.  SOLIUM.     By   i;  r$e  of  a  proposition  is  meant  a 

proposition  which  has  the  hypothesis  of  the  first  as  conclusion 
and  the  conclusion  of  the  first  as  hypothesis.  The  converse  of 
a  truth  is  not  neceuarilg  true.     Thus,  parallel  lines  never  meet ; 

r  ?/*e^  are  parallel,  is  not  true  unless 
the  lines  lie  in  the  same  plane. 


v..  The  <<  >n  verse  of  many  propositions  will  be  omitted, 
but  their  statement  and  demonstration  shomel  be  required  as  an 
import  rise  for  the  student. 


28  GEOMETRY. BOOK    I. 


Proposition  XII.     Theorem. 

69.  Conversely  :    When  two  straight  lines  are  cut  by  a 

third  straight  line,  if  the  alternate-interior  angles  he  equal, 
the  two  straight  lines  are  parallel. 


Let  E  F  cut  the  straight  lines  A  B  and  C  I)  in  the  points 
II  and  K,  and  let  the  Z  A  II K  =  Z  II KB. 

We  are  to  }~>rove      A  B  II  to  C  D. 
Through  the  point  II  draw  MN  II  to  CD; 

thm  Z  MIIK=Z  I/KIJ,  §68 

(being  alt. -int.  A  ). 

Bui  Z  AIIK  =  Z  II K  I),  Hyp. 

.'.Z  M II  K  =  Z  A  UK.  Ax.  1. 

.*.  the  lines  M  N  and  A  B  coincide. 
But  J/iVis  II  to  CD;  Cons. 

.".  A  B,  which  coincides  with  M  N,  is  II  to  C  I>. 

Q.  E.  D. 


PABALLEL    I.IXES.  29 


Proposition  XIII.     Theorem. 

70.    If  two  parallel  lines  be  cut  by  a  third  straight  line, 
ike  '  Interior  angle*  are  equal. 

E 


Let  A  B  and  CD   be   two  parallel  lines    cut   by  the 
straight  line  E  /',  in  the  points  II  and  K. 

H  v       /  EU  H  =  Z  II KD. 

Z  EJIIi  =  ZAII  K\  §49 

A). 

But  ZAIIK=Z1IA  />.  §68 

-.-////.  A). 

.-.Z  E II  Jl  =  Z  H  KD.  Ax.  1 

In  like  manner  we  may  prove 

ZEIIA=ZIIKC. 

Q.  E.  D. 


71.   CoBOLLARY.    The  alternate-exterior  angles,  E II B  and 
( '  K  I\  and  also  A  II  E  and  D  K  E,  are  equal. 


30  GEOMETRY.  —  BOOK   I. 


Proposition  XIV.     Theorem. 

72.  Conversely  :  When  two  straight  lines  are  cut  by  a 
third  straight  line,  if  the  exterior-interior  angles  be  equal, 
these  two  straight  lines  are  parallel. 


E 


Let  E  F  cut    the  straight  lines  AB  and  CD  in   the 
points  II  and  Kf  and  let  the  Z  K II B  =  Z  UK/). 

Wt  are  to  prove      A  B  II   to  C  D  . 
Through  the  point  II  draw  the  straight  Line  M  X  II  to  CD. 

Then  Z  KIIN=Z  II  K  IK  §  7C 

ng  r.rt.-'nil.  A). 

But  Z  KIIB  =  Z  UK  D.  Hyp 

.-.  Z  EHB  -  Z  EHN.  Ax.  1 

.*.  the  lines  MX  and  A  B  coincide. 
Bui  MX  is  II  to  CD,  Cons 

.'.  A  B,  which  coincides  with  M JX,  is  II  to  CD. 

Q.  E.  D. 


PABALLEL    LINES.  31 


Proposition  XV.     Theorem. 

73.  If  two  parallel  lines  be  cut  by  a  third  straight  line, 
He  turn  of  (//>'  too  inferior  angles  on   the  same  side  of  the 
>f  H ne  is  equal  (<>  two  fight 


Let  All  and  C  D  be    two  parallel   lines    cut   by   the 
straight  line   EF  in  the  points  II  and  K 

We  a  m      Z  B  UK  +  Z  UK  I)  =  two  rt.  A. 

Z  EUB  +  Z  BIIK=2  it.  A,  §  34 

A). 

Bui  Z  Ell B  =  Z  UK  lh  §  70 

'■"j  c.rt.-int.  A). 
gtitate  Z  II  K  I)  for  Z  #7/7?   in  the  first  equality; 
then  Z  B II K  +  Z  IIKO  =  2  rt.  A 

Q.  E.  D. 


32  GEOMETRY. BOOK   I. 

Proposition  XYI.     Theorem. 

74.  Conversely  :  When  two  straight  lines  are  cut  o//  a 
third  straight  line,  if  the  two  interior  angles  on  the  same  side 
of  the  secant  line  be  together  equal  to  tivo  right  angles,  then 
the  two  straight  lines  are  parallel. 

E 


Let  EF  cut  the  straight  lines  AB  and  CD  in  the 
points  II  and  K,  and  let  the  Z  B  II  K  +  Z  II K D 
equal  two  right  angles. 

W(  "re  to  prove      AB  II  to  C  D. 

Through  fche  point  II  draw  UN  II  to  CD. 

Then  Z  NIIK  +  Z  II K  D  =  2  rt.  A,  §  73 

(h  ing  two  interior  A  on  f/ir  some  side  of  the  secant  line). 

But  Z  B1IK+  Z  II Kl)  =  2  rt.  A.  I [yp. 

.'.ZX1IK+  Z  II  KI)  =  ZB1IK+  ZII K  I).     Ax.  1. 

Take  away  from  eacli  of  these  equals  the  common  Z  II  K  IK 

then  Z  NHK=  Z  B II  K. 

.'.  the  lines  A  B  and  M N  coincide 

But  MNvt  II  to  CD;  Cons. 

.*.  A  B,  which  coincides  with  M N,  is  II  to  CD. 

Q.  E    D. 


PARALLEL    LINES. 


33 


Proposition  XVII.     Theorem. 

7.").    Two  straight   lines  which  are  parallel  to  a  third 
Uraighi  line  are  parallel  to  each  other. 


A- 

C- 


B 
-D 

F 


Let  A  B  and  C  D  be  parallel  to  E F. 
We  are  toprem     A  B  II  to  CD. 

Draw  //A'_L  to  EF. 
Since  CD  and  EFm  1,  ffJTis  J.  to  CD,         §  67 

gki  /hie  be  J_  to  one  of  two  fb,  it  is  ±  to  the  other  also). 

Since  A  B  and  BF*te  II,  UK  is  also  ±toAB,      §  67 
.'.Z  IIOB  =  Z  II PD, 

{each  being  art.  Z). 

.'.  A  B  is  II  to  CD,  §  72 

we  cut  by  a  third  straight  line,  if  tlie  cxt.-int.  A 
qnalf  the  two  lines  are  II  ). 

Q.  E.  D. 


34 


GEOMETRY.  • 


■  BOOK   I. 


Proposition  XVIII.     Theorem. 

76.   Two  parallel  lines  are  everywhere  equally  distant 

from  each  other. 

i:                 M                II 
A 1 1 ; B 


Let  A  B  and  CD  be  two  parallel  lines,  and  from  any 
two  points  in  A  B,  as  E  and  II,  let  EF  and  UK 
be  drawn  perpendicular  to  A  I>. 
We  ore  t<>  prove      E  F  ■=  //  K. 

Now  EF  and  //  K  are  i.  to  C  h,  §  G7 

{>'  live  _L  t<>  n„,  of  two  \\s  is  _L  to  Ho  other  also). 

Let  M  he  the  middle  point  of  EH. 
Draw  MP±  to  A  B. 
On  MP  as  an  axis,  fold  over  the  portion  of  the  figure  on 
the  right  of  MP  until  it  eoraee  into  the  plane  of  the  figure  on 
the  left. 

M B will  fall  on  MA, 
{for  Z  P  M 11=  ZP  M  /•:,  i  ach  being  a  rt  Z )  ; 

the  point  //  will  fall  on  E, 
(forMH=  ME,  by  hyp.)  ; 

If  K  will  fall  on  EF, 

{for  Z  M JIK=  Z  M E F,  each  being  a  rt.  Z)  ; 

and  the  point  K  will  fall  on  E  E,  or  E F  produced. 

Also,  PD  will  fall  on  PC, 
(Z  MPK=  Z  MPF,  earlt  being  a  rt.  Z)  ; 

and  the  point  K  will  fall  on  P  C. 

Since  the  point  K  fills  in  both  the  lines  EFmd  PC, 

it  must  fall  at  their  point  of  intersection  E. 

.'./IE=EF,  §18 

(their  'xtremities  being  the  same  points). 

Q.  E.  D. 


PARALLEL    LINES. 


35 


Proposition  XIX.     Theorem. 

77.    Two  mi i } lee  whose  sides  are  parallel,  two  and  two, 
Hud  lie  in  ike  name  direction^  or  opposite  directions,  from  their 
■t'S,  are  equal, 
A         D 


D> 

Pig.  l. 

Let  A  11  and  E  (Fig.  1)  have  their  sides  BA  and  ED, 
and  BO  and  BF  respectively,  parallel  and  lying 
in  the  same  direction  from  their  vertices. 

(Pi  an  to  pro*  the      A  B  =  A  B, 

ay)  two  ridea  which  an.*  not  II  until  they 
intersect,  as  at  //  ; 

A  n  =  ZDHC,  §  70 

(beii  A), 

and  AE  =  AD1I<\  §70 

.-.z  n  =  z  /:.  Ax.  l 

Let  A  B  and  K'  (Fig.  2)  have  B'  A'  and  E'  /)',  and  B'  C 
and  E'  F  respectively,  parallel  and  lying  in  oppo- 
site directions  from  their  vertices. 
PPi  on  /  Z  B'  =  Z  E'. 

Produce  (if  necessary)  two  sides  which  are  not  II  until  they 
intersect,  as  at  II'. 

Th  ABt  =  AE'irC'f  §70 

(being  txL-inL  A), 

and  A  E'  =  A  E'lI'C,  §  08 

(Jbeing  alt. -int.  A)  ; 

.-.  A  B'  =  A  E',  Ax.  1. 

Q.  E.  D. 


oG  GEOMETRY. BOOK    I. 

Proposition  XX.     Theorem. 

78.  If  tiro  angle*  have  two  sides  parallel  and  lying  in 

the  same  direction  from  their  vertices,  while  the  other  two 
sides  arc  parallel  and  He  in  opposite  directions,  then  the  two 
a, i  pies  are  supplements  of  each  other. 

c 


Let  A  EC  and  1)  E  F  be  two  angles  having  BC  and  E  I) 
parallel  and  lying  in  the  same  direction  from  their 
vertices,  while  EF  and  B  A  are  parallel  and  lie  in 
opposite  directions. 

W\  on  to  prove  Z  ABC  and  Z  D E F  supplements  of  each 

Produce  (if  necessary )  two  sides  which  ate  not  II  until  they 
Intel  //. 

Z  A  BC  =  Z  BUD,  §  70 

{being  aot.-int.  A). 

z  i)  ef=z  BB&t  §  68 

(h,  fag  ,f  11. -int.  A). 

Bat  Z  n  If  I)  and  Z  11  If  E  are  supplements  of  each  other,    §  3 t 
(being  sup. -adj.  A ). 

.-.  Z  A  BO  and  Z  I)  EF,  the  equals  of  Z  BUD  and 
Z  B  ll  /;.  are  supplements  <>f<-;trii  other. 

Q.  E.  D. 


TRIANGLES.  37 


On  Triangles. 

79.  Def.    A  Triangle  is  a  plane  figure  bounded  by  three 
straight  lines. 

A  triangle  has  six  parts,  three  sides  and  three  angles. 

80.  When  lli.-  rix   parts  of  one  triangle  are  equal  to  the  six 
parte  of  another  h  ich  to  each,  the  triangles  are  said  to 

><  <ill  respects. 

81.  Def.    In  two  equal  triangles,  the  equal  angles  are  called 
tfomol  s,  and  the  equal  sides  are  called  Homologous 

82.  In    equal    triangles   the    equal    sides  are   opposite   the 
equal  angles. 


EQUILATERAL. 


83.   I  Mr.    A  Seal  _de  is  one  of  which  no  two  sides 

(iial. 

>l.    I  > i  i .    An   Tmscdes  triangle  is  one  of  which  two  sides 

mal. 

I  >i:i  .    An  Equilateral  triangle  is  one  of  which  the  three 
sides  are  equal 

I'll.    Tin-   Base  of  a  triangle  is  the  side  on  which  the 
triangle  is  supposed  to  stand. 

In  an  isosceles  triangle,  the  side  which  is  not  one  of  the 
equal  sides  is  considered  the  base. 


38 


-UOOK    I. 


87.  Def.    A  :  angle  is  one  which  has  one  of  the 

_  ht  angle. 

88.  DBF.     The  side  opposite  the  right  angle  is  called  the 
<>tenu*e. 

89.  \h.\ .    An   Obtuse  triangle  is  one  which  has  one  of  the 
I  an  olAu.se  angle. 

90.  1  h.v.    An  Acute  triangle  is  one  which  has  all  the  angle* 


CQUIANOULAK. 


91.    Ih.v.    A  f/ular  triangle  is  one  which  lias  all 

Def.    In  anv  to  the  hase  is 

le,  and  its  vertex  is  called  I 

i  -in  the  vertex  to  the  haw-,  or  the  bote  produced. 
Def.    Tl 
eluded  between  a  side  and  an  adjacen  /  CB  h. 

Def.   The  two  angle*  of  a  Mangle  which  are  opp 
.  are  called  the  tw< 
I  sad  C. 


TRIANGLES.  39 


96.  Any  side  of  a  triangle  is  less  than  the  sum  of  the 
other  two  sides. 

e  a  straight  line  is  the  shortest  distance  between  two ' 

points, 

A  C  <  A  B  +  /; 

97.  Any  side  of  a  triangle  is  greater  than  the  difference 
of  the  other  two  sides. 

In  the  inequality  A  C  <  A  B  +  BC, 

take  away  A  B  from  each  side  of  the  inequal 

Then  -AB<BC;  or 

BO  AC- A  B. 


-  now  that  the  sum  of  the  distances  of  any  point  in  a 
triangle  from  the  vertices  of  three  angles  of  the  triangle  is  greater 
than  half  the  sum  of  the  sides  of  the  triangle. 

v  that  the  locus  of  all  the  points  at  a  given  distance 
from  a  given  straight  line  A  B  consists  of  two  parallel  lines, 
drawn  on  opposite  sides  of  A  B,  and  at  the  given  distance 
from  it 

3.  Show  that  the  two  equal  straight  lines  drawn  from  a  point 

.t  line  make  equal  acute  angles  with  that  line. 

4.  Show  that,  if  two  angles  have  their  sides  perpendicular, 
each  to  each,  they  are  either  equal  or  supplementary. 


40  GEOMETRY. —  BOOK    I. 


Proposition  XXI.     Theorem. 

98.  The  sum  of  the  three  angles  of  a  triangle  is  equal 
to  two  right  angles. 

B  E 


* gr r 

Let  A  B  C  be  a  triangle. 

IT  on  to  prove    Z  B  Sr  Z  B  C  A  +  Z  A  =  two  rt.  A. 

Draw  C  E  II  to  A  B,  and  prolong  A  C. 

Then  Z  BCF  +  Z  ECB  +  Z  BCA  =  2  rt.  A,        §  34 
turn  of  all  the  A  about  a  point  on  '■  'de  of  a  straight  line 

=  2  rt.  A  ). 

Bat  ZA=ZECF,  §70 

i»  i/i'f  ,,t..;, a.  A), 

and  Z  B  =  Z  BCE,  §  68 

(being  a  If. -int.  A  ). 

Substitute  fatZ  EC  Fund  Z  BCEtheia equals,  A  and  ft 

Then        Z  A  +  Z  n  +  Z  BCA  =  2  rt.  A. 

Q.  E.  D. 

99.  Corollary  1.  If  the  sum  of  two  angles  of  q  triangle  be 
known,  the  third  angle  can  be  found  by  taking  this  sum  from 
two  right  angles. 

LOO.  Cor.  2.  If  two  triangles  have  two  angles  of  the  one 
equal  to  two  angles  of  the  other,  the  third  angles  will  be  equal 


TRIANGLES.  41 


101.  Cor.   3.    If  two  right  triangles  have  an  acute  angle 
of  thf  one  equal  to  an  acute  angle  of  the  other,  the  other  acute 

will  be  equaL 

102.  (  Job.  1.   In  a  triangle  there  can  he  hut  one  right  angle, 
obtuse  an 

103.  Cor.  5.    In  a  right  triangle  the  two  acute  angles  are 
complements  of  each  oth 

1<>1.  (Job.  6,    In  an  equiangular  triangle,  each  angle  is  one 
third  of  two  right  an  w<_>  thirds  of  one  right  angle. 


PROl  XXII.       TlII.oKKM. 

105.    /  i  r'miKj  It  is  equal  fo  Ih  e  8  u  ui 

'>/  thr  //<■<  tUtf/feS. 


a 

Let  BC  If  be  an  exterior  angle  of  the  triangle  ABC. 
II,  em  to  j'mve      Z  BCIf=Z  A  +  Z  B. 

Z  BC  11+  Z  A  C  B  =  2  it.  A,  §  34 

A). 

Z  A  +  Z  B  +  ZACB  =  2  rt.  A,  §  98 

(three  AofaA  =  two  rt.  A  ). 

.'.ZBCH+,  ^ZA  +  ZB  +  ZACB.      Ax.  1. 

Take  away  from  each  of  these  equals  the  common  ZA  CB\ 
then  Z  BC  11  =  Z  A  +  Z  B. 

Q.  E.  D. 


42 


GEOMETRY. 


BOOK   I. 


Proposition  XXIII.     Theorem. 

106.  Two  triangles  are  equal  in  all  resjjects  when  two 
sides  and  the  included  angle  of  the  one  are  equal  respectively 
to  two  sides  and  the  included  auyle  of  the  other. 


B  At 


In    the    triangles   ABC   and  A' £' C,   let  AB  =  A'B', 
A  C=A'C',A  A=Z.  A'. 

We  are  to  prove      A  A  B  C  =  A  A'  B'  C. 

Tab- 1  up  the  A  A  BC  and  place  it  upon  the  A  J'/j'C  so 
that  A  B  shall  coincide  with  A'  B'. 


Then        A  C  will  take  the  direction  of  A1  C, 
(for  /.  A  =  A  A',  by  /<!/!>.), 

the  point  C  will  fall  upon  tin'  point  O, 
(forAC=A'C'y  by  hyp.)  ; 

.'.  CB  =  C  B', 
(their  extremities  being  the  same,  points). 

•\  the  two  A  coincide,  and  are  equal  in  all  respects. 


§  18 


Q.  E.  D. 


TRIANGLES.  43 


Proposition  XXIV.     Theorem. 

107.  Tiro  triangle*  are  equal  in  all  respects  when  a  side 
and  two  adjacent  angles  of  the  one  are  equal  respectively  to  a 
side  and  two  adjacent  angles  of  ike  other. 

c  a 


n  ,v 


In    the   triangles    ABC   and  A'  B' C,    let  A  B  =  A'  B'y 
Z  A=Z  A',  A  B  =  Z.  B'. 

We  are  to  prove      A  A  BC  =  A  A'  B'  O. 

Take  up  A  A  BC  and    place   it   upon  A  A'  B' C,   so   that 

A  B  shall  •  oiacide  with  At  l>'. 

A  C  will  take  the  direction  of  A1  C, 
(/or  Z  A  =  Z  A',  by  hyp.)  ; 

the  poinl  C.  the  extremity  of  A  6',  will  fall  upon  4' C"  or 
A'  C  prodaced 

B  C  will  take  the  direction  oiB'C, 
(forZB  =  ZBf,  by  hyp.); 

the  poinl  C.  the  extremity  of  BC,  will  fall  upon  B' C  or 
B1  C  prodn 

.'.  the  point  C,  falling  upon  both  the  lines  A'  C  and  B1  C, 
must  fall  apon  a  point  common  to  the  two  lines,  namely,  C. 

.'.  the  two  A  coincide,  and  are  equal  in  all  respects. 

Q.  E.  D. 


44 


GEOMETRY.  BOOK    I. 


Proposition    XXV.      Theorem. 

108.  Tiro  triangles  are  equal  when  the  three  sides  of  the 
qne  are  equal  respectively  to  the  three  sides  of  the  other, 
B  B> 


In    the   triangles   A  B  C   and  A'  B'  C,   let  A  B  =  A'  B', 
AC  =  A'C,  BC  =  B' C 
in  on  to  prove      A  A  BC  =  A  A' B' C. 
Place  A  A'  B'  C  in  the  position  A  B'  C,  haying  its  greatest 
side  A' C  in  coincidence  with  its  equal  AC,  and  its  vertex  at 
ll1,  opposite  B. 

Draw  B  B'  intersecting  A  C  at  //. 

Snu-At  AB  =  A  /:\  Hyp. 

point  A  is  at  equA]  distances  bom  B  and  B'. 

Since  J?  C  =  /?7  C,  Hyp. 

point  6'  La  at  equal  distances  from  B  and  B1. 

.'.  A  C  is  J_  to  J5  J5'  at  its  middle  point,  §  60 

(two  points  at  equal  distant  i  from  the  extremities  of  a  straight  line  deter' 

"  the  J.  at  the  middle  of  that  line). 

Now  [f  A  A  B'  C  be  folded  over  on  .4  (7  as  an  axis  until  it 
comes  into  the  plane  of  A  ABC, 

II  H'  will  tall  on  EB} 
{for  Z  A  H  B  =  Z  A  J I  /;',  each  h  ing  ><  rt,  Z), 

and  point  B1  will  fall  on  B, 
{for  H  B'  =  H  B). 

.'.  the  two  A  coincide,  and  are  equal  in  all  respects. 

Q.  E.  D. 


TRIAXGLES. 


45 


Proposition   XXVI.      Theorem. 

L09.  Two  right  triangles  are  equal  when  a  side  and  the 
hypotenuse  of  the  one  are  equal  respectively  to  a  side  and  the 
hypotenuse  of  the  other. 


In  the  right  triangles  ABC  and  A'  B'  C,  let  A  B  =  A'  5', 
and  AC  =  A'C. 

We  are  to  prove      A  A  BC  =  A  A'  B'  C. 

Tftkfl  up  the  A  A  BC  and  place  it  upon  A  A' B' C,  so  that 
A  B  will  oomdde  with  A' B'. 

Then  BC  will  fell  upon  n'C, 

(forZABC=ZA'B'C  \g a  rt.  Z), 

and  point  C  will  fall  upon  C ; 

otherwise  the  equal  oblique  lines  A  (7- and  A'  C  would  cut 
off  unequal  distances  from  the  foot  of  the  JL,  which  is  im- 
possible, §  57 

((>/•<>  (qua  J  sfrom  a  point  in  a  X  cut  off  equal  distances  from  the 

foot  of  the  _L  ). 

.*.  the.  two  A  coincide,  and  are  equal  in  all  respects. 

Q.  E.  D. 


46 


GEOMETRY. BOOK    I. 


Proposition   XXVII.      Theorem. 

110.  Two  right  triangles  are  equal  when  the  hypotenuse 
and  an  acute  angle  of  the  one  are  equal  respectively  to  the 
hypotenuse  and  an  acute  angle  of  the  other. 


In  the  right  triangles  ABC  and  A'  B'  C,  let  AC  =  A'  C, 
and  Z  A=  Z  A'. 


We  are  to  prove      A  A  B  C  =  A  A'  B'  <?'. 
A  C  =  A'  C, 
Z  A=Z  A>, 


Hyp. 
Hyp. 


then  Z  C  =  Z  C,  §  101 

100  rt.  A  have  an  acute  Z  of  the  one  equal  to  an  acute  Z  of  the  other, 
then  the  other  acute  A  are  equal). 

.'.AABC  =  AA'B'C,  §  107 

(two  &  are  equal  when  a  side  and  two  adj.   A  of  the  one  are  equal 
respectively  to  a  side  and  two  adj.  A  of  the  other), 

Q.  E.  D. 

111.  Corollary.  Two  right  triangles  are  equal  when  a 
side  and  an  aeute  angle  of  the  one  are  equal  respectively  to  an 
homologous  side  and  acute  angle  of  the  other. 


TRIANGLES.  47 


Proposition  XXVIII.   Theorem. 
112.    In   nit   uoteete*  triangle  the  angles  opposite  the 

c 


I  B 

Let  ABC  be  an  isosceles  triangle,  having  the  sides 
A  C  and  C  B  equal. 

We  are  to  prove      Z  A  =  Z  B. 

From  C  draw  the  straight  line  C  E  so  as  to  bisect  the 
ZACB. 

In  the  A  it  CA'an.l  BOB, 

AC=BC,  Hyp. 

C  E=C  Ef  Iden. 

ZACE=ZBCE;  Cons. 

..AACE  =  ABCE,  §106 

-  sides  and  the  included  Z.  of  the  one  are  equal 
respectively  to  two  sides  and  the  included  Z.  of  the  other). 


r.Z  A=Z  B, 

(being  homologous  A  of  equal  A  ). 


Q.  E.  D. 


If  the  equal  sides  of  an  isosceles  triangle  be  produced, 
show  that  the  angles  formed  with  the  base  by  the  sides  produced 

'[Ual. 


48 


GEOMETRY.  —  BOOK    I. 


Proposition  XXIX.     Theorem. 

113.  A  straight  line  which  bisects  the  angle  at  the  vertex 
of  an  /'.v.y -r/r.v  triangle  divides  the  triangle  into  tiro  equal 
triangles,  is perpendi'-ular  to  the  base,  and  bisert.s  the  base. 

C 


Let  the  line  C  E  bisect  the  Z  AC  B  of  the  isosceles 
A  ACB. 

Wi  em  to  prove      I.  AACE  =  ABCE; 

II.  line  CE±toAB; 
III.  A  E  =  BE. 

I.    In  thoA  ACE  ai»\  BCE, 

AC=BC,  Hyp. 

CE=C/.\  Iden. 

Z  ACE  =  Z  BCE.  Cons. 

.-.A  AC  E  =  A  BCE,  §  10G 

ag  two  sides  and  /  ,fual  respectively  to  two  aides 

and  the  included  A  of  the  otJier). 

Also,  II.  Z  CEA  =  Z  CEB, 

{being  homologous  A  of  equal  &). 

.*.  CEis±  to  AB, 
(a  straight  line  meeting  another,   making  tlie  adjacent  A  equal,  is  ±  to 

that  line). 

Also,  III.  AE=  E  /;, 

(being  homologous  sides  of  equal  &). 

Q.  E.  D. 


TRIANGLES.  49 


Proposition  XXX.     Theorem. 

114.    If  Udo  angle*  of  a  triangle  be  equal,  the  sides  op- 
lr  the  equal  angles  are  equal,  and  the  triangle  is  isosceles. 


B  D 

In  the  triangle  ABC,  let  the  Z  B  =  Z  C. 
Wi  ve      A  B  =  A  C. 

Draw  A  D  ±  to  B  C. 
In  the  it.  A  A  DBund  A  l><\ 

A  I>  -    A  Dt  Iden. 

ZB  =  ZC, 

.'.  rt.  A  A  1)  B  =  rt.  A  A  DC,  §  111 

>g  a  side  and  an  acute  /L  oj  !>>al  respectively  to  a  side  and  an 

Z.  of  the  "I 

:.AB  =  AC, 
(being  homologous  sides  of  equal  &). 

Q.  E.  D. 


Ex.    Show  that  an  equiangular  triangle  is  also  equilateral. 


50  GEOMETRY. BOOK   I. 

Proposition  XXXI.     Theorem. 

115.  If  two  triangles  have  two  sides  of  the  one  equal 
respectively  to  two  sides  of  the  other,  but  the  included  angle 
of  the  first  greater  than  the  included  angle  of  the  second,  then 
the  third  side  of  the  first  will  be  greater  than  the  third  side 
of  the  second. 


C 

\\l 

E  *1e 

In  the  A  ABC  and  ABE,  let  A  B  =  A  B,   B  C  =  B  E ; 
but  Z  ABO  Z  ABE. 

We  are  to  prove     AO  A  E. 
Place  the  A  so  that  A  B  of  the  one  shall  coincide  with  A  B 
<>f  the  other. 

Draw  BFaotBto  bisect  Z  EBC. 
Draw  EF. 
In  the  a  BBFaad  OBF 

i:  B  =  BC,  Hyp. 

BF=BF,  Men. 

Z  EBF=Z  CBF,  Cons. 

.'.  the  A  EBFund  C B F  are  equal,  §  10G 

{hnrng  two  sides  and  the  included  £  of  one  equal  respectively  to  two  sides 
and  the  included  Z.  of  the  other). 
.\EF=FC, 
(being  homologous  tides  of  equal  A). 
Now  A  F  +  F  3  >  A  Bf  §  96 

(the  sum  of  two  sides  of  a  A  is  greater  than  the  third  side). 
Substitute  for  F  E  its  equal  FC.     Then 
AF+  FOAE;  or, 
A  C  >  A  E. 

Q.  E.  D. 


TRIANGLES.  51 


Proposition  XXXII.     Theorem. 

116.  Conversely:  If  two  sides  of  a  triangle  be  equal 
respectively  to  two  sides  of  another,  but  the  third  side  of  the 
first  triangle  be  greater  than  tie  third  side  of  the  second,  then 
tlir  angle  opposite  tie  third  side  of  the  first  triangle  is  greater 
tin m  tie  angle  opposite  the  third  side  of  the  second. 


In  the  &ABCandA'B'C'J  let  A  B  =  A'  B',  AC  —  A'  C ; 
but  HO  B'  C. 

We  are  to  prove      Z  A  >  Z  A'. 
If  Z  A=Z  A', 

then  would       AABC  =  AAfB'C,  §106 

(having  two  sides  and  the  included  Z  oftlie  one  equal  respectively  to  two  sides 
and  (  I  £of  tlie  other), 

and  BC^B'C, 

(being  homologous  sides  of  equal  & ). 

And  if  A  <  A', 

thru  would  BC<B'C,  §115 

(if  two  sides  of  a  Abe  equal  respectively  to  two  sides  of  another  A,  but  the 
4  be  greater  than  the  included  Z  of  the  second,  the 
will  be  greater  than  the  third  side  of  the  second.) 

But  both  these  conclusions  are  contrary  to  the  hypothesis; 

.*.  Z  A  doea  not  equal  Z  A',  and  is  not  less  than  Z  A'. 

:.Z  A>  Z  A1. 

Q.  E.  D 


52  GEOMETRY.  —  BOOK    I. 


Proposition  XXXIII.     Theorem. 

117.    Of  two    sides  of  a   triangle]   that  is  the  greater 
which   is  opposife  the  greater  angle. 


In  the  triangle  ABC  let  angle  AC B  be  greater  than 
angle  /:. 

Wi  we      A  B  >  A  C. 

Draw  CEbq  aa  bo  make  Z  B  C  E  =  Z  B. 

Then  EC  =  EB,  §114 

A ). 

Now  AE+  EOAC,  §96 

(/A-  wm  of  two  tides  of  a  A  /*■  greater  than  the  third  side). 

t  i t  ute  for  i£  C  its  equal  .#  A     Then 

ii  ^  +  E  B>  AC,  or 

AB>  A  C 

Q.  E.  D. 


Ex.  ABC  and  ABB  are  two  triangles  on  the  same  base 
^1  j5,  and  on  the  same  side  of  it,  the  vertex  of  each  triangle 
being  without  the  other.  If  A  C  equal  A  D,  show  that  B  C 
cannot  equal  B  D. 


TRIANGLES.  53 


Proposition   XXXIY.     Theorem. 

lis.   Of  two  angle*  of  a  triangle,  that  is  the  greater 
which  is  opposite  the  greater  side. 

B 


C 

In  the  triangle  A  BC  let  A  B  be  greater  than    I  C. 
PPi  tm  to  prove      Z  A  C  B  >  Z  /;. 

Tak  ,ual  to  A  C; 

Draw  EC. 

Z  AEC  =  Z  ACE,  §  112 

i 

But  Z  A  EC>  Z  /;,  §  105 

(an  exterior  A  of  a  A  is  gr  Z. ), 

an. I  Z  AC B  >  Z  ACE. 

Substitute  foi  Z  A  C  E  its  equal  Z  A  EC,  then 

Z  ACB>  Z  A  EC. 

h  mure  is  Z  AC  B>  Z  B. 

Q.  E.  D. 


If  the  angles  ABC  and  ACB,   at  the  base  of  an 
be  bisected  by  the  straight  lines  B D,  CD, 
Bhow  that  D BC  will  be  an  isosceles  triangle. 


54  GEOMETRY. BOOK   I. 


Proposition  XXXY.     Theorem. 
119.   The  three  bisectors  of  the  three  angles  of  a  triangle 
m •trt  in  a  point. 


Let  the  two  bisectors  of  the   angles  A  and  C  meet 
at  0,  and  0  B  be  drawn. 

xm      B  I  he  Z  B. 

Draw  the  J8  OK,  ()  />,  and  OH. 

CP, 
OC=OC,  Iden, 

Z  OCK  =  Z  OCP,  (lis. 

.'.A  OCK  =  A  OCP,  §  110 

M  <mtf  an  ac^fe  Z  «/'  parf  respectively  to  the 

(ho)n  >'■  i  ofeq/kal  A). 

In  the  it,  A  Oil  P  and  O  J  //, 

0A*=OAt  Lien. 

-ZOAP  =  ZOAH,       ,  Cona. 

/.A  6>d  P  =  AOA  //,  §  110 

w  /A-'  KypoUnxm  and  cm  acute  Z  of  the  one  eq ""I  respectively  to  the 
hypotenuse  and  an  acute  Z  of  the  0th 

.'.  OP=0/t. 
(being  homologous  tides  "/equal  A  ). 

But  we  have  already  shown  0  P  =  O  A\ 

.'.OH=()k\  Ax.  I 

Now  in  rt.  A  0  II  B  and  0  KB 


TRIANGLES.  55 


Oil  =0  K,  a.ndOB  =  OB, 

.'.AOHB  =  A  OKB,  §  109 

of  the  one  equal  respectively  to  the  hypote- 


nuse and  a  side  of  the  other), 

.-.  Z  o/;//  =  Z  OBK, 
(being  homologous  A  of  equal  A  ). 


Q.  E.  D. 


Proposition   XXXVI.     Theorem. 
120.    The   i  peudicular*  erected  at  the  middle 

point*  qf  the  thn  yf  a  triangle  meet  in  a  point. 

A 


F 

Let  I)  D ,    /   /  ,    /'/',   be  three  perpendiculars  erected 
at  I),  E,  F,  the  middle  points  of  A  B,  A  C,  and  BO. 

We  are  to  pro*  iat,  as  0. 

The  two  J*  I)  D'  and  BE1  meet,  otherwise  they  would  be 
parallel,  and  A  B  and  A  £7, being  _l!  to  these  lines  from  the  same 
point  A,  would  he  in  the  same  straight  line; 

but  this  is  impossible,  since  they  are  sides  of  a  A. 

Let  0  be  the  point  at  which  they  meet. 

Then,  since  0  is  in  J)  D',  which  is  J_  to  A  B  at  its  middle 

point,  it  is  equally  distant  from  A  and  B.  §  59 

Also,  since  0  is  in*  E  E' ,  _L  to  A  0  at  its  middle  point,  it  is 
equally  distant  from  A  and  0.  / 

.*.  0  is  equally  distant  from  B  and  C ; 

.'.  0  is  in  F F'  _L  to  BO  at  its  middle  point,      .    §  59 

(the  locus  o)  \ly  distant  from  the  extremities  of  a  straight  line 

is  the  ±  erected  at  the  middle  of  that  line). 

Q.  E.  D. 


56 


GEOMETRY. BOOK    I. 


Proposition  XXXYII.     Theorem. 

1:21.   The  three  perpendiculars  from  the  vertices  of  a  tri- 
angle to  the  opposite  sides  meet  in  a  point. 

B 


In  the  triangle  ABC,  let  B  P,  A  II.  OK,  be  the  per- 
pendiculars from  the  vertices  to  the  opposite 
sides. 

We  are  to  prove  they  meet  in  some  point,  as  0. 

Through  the  vertices  A,  B,  C,  draw 

A'  B  II-  to  BC, 

A'  C  II  to  A  C, 

B'  C  II  to  A  B. 

In  the  A  A  B  A'  and  A  B  C,  we  have 

A  B  =  A  B, 


ZABA'  =  Z  BAC, 

(being  alternate  interior  A  ), 

Z  BAA'  =  Z  ABC. 


Iden. 
§  68 

§  68 

.*.A  ABA'  =  A  ABC,  §  107 

Hide  and  two  adj.  A  of  tlie  one  equal  respectively  to  a  side  mm 
two  adj.  A  of  the  other). 

.'.A'B  =  AC\ 

(being  homologous  sides  of  egical  A  ). 


TIM  57 


In  the  A  CBC  mid  A  B( 

BC  =  BC,  Idea 

ZCBC'  =  ZBCA,  §08 

(being  alternate  interior  A ). 

ZBCC'  =  ZCBA  §68 

.-.A  CBL"  =  AABC,  §107 

nrj  a  side  o>  A  of  the  one  equal  respectively  to  a  M 

adj.  A  of  the  other). 

BC'  =  AC, 
{being  homologous  sides  of  equal  A ). 

But  we  have  already  shown  A'  B  =  A  C, 

A'B  =  BC,  Ax.  1. 

/.Bjb\  We  point  of  A'C. 

BPia±toAC,  Hyp. 

\s±toAU  §  67 

(a  straight  line  which  is  ±  to  one  of  two  \\s  is  ±  ( 

middle  point  of  A'  C  \ 

.-.  BP  is  A.  to  A'C  at  its  middle  point 

In  like  manner  we  may  prove  tl 

A  H  i>  _L  t<»  A'  W  at  its  middle  point, 

and  C  K  A.  to  B1  C  at  its  middle  point. 

.-.  11  l\  A  //.   and   CK  are  Js  erect-  middle  points 

.  .  A    B'C. 

.'.  these  J§  meet  in  a  point.  §  120 

points  of  the  sides  of  a  A  ///  >int). 

Q.  E.  D. 


58 


GEOMETRY. BOOK    I. 


On  Quadrilaterals. 

122.  Def.    A  Quadrilateral  is  a  plane  figure  bounded  by 
four  straight  lines. 

123.  Def.    A  Trapezium  is  a  quadrilateral  which  has  no 
two  sides  parallel. 

124.  Def.    A  Trapezoid  is  a  quadrilateral  which  has  two 
sides  parallel. 

125.  Def.    A  Parallelogram  is  a  quadrilateral  which  has 
its  opposite  sides  parallel. 


TRAPEZIUM. 


TRAPEZOID. 


PARALLELOGRAM. 


126.  Def.    A  Rectangle  is  a  parallelogram  which  has  its 
angles  right  angles. 

127.  Def.    A   Square   is   a    parallelogram   which   has   its 
angles  right  angles,  and  its  sides  equal. 

128.  Def.    A  Rhombus  is  a  parallelogram  which  has  its 
sides  equal,  but  its  angles  oblique  angles. 

129.  Def.    A  Rhomboid  is  a  parallelogram  which  has  its 
angles  oblique  angles. 

The  figure  marked  parallelogram  is  also  a  rhomboid. 


RECTANQLE. 


QUADRILATERALS.  59 


130.  I'i.f.  The  side  upon  which  a  parallelogram  stands, 
tnd  the  opposite  side,  are  called  its  lower  and  upper  bases;  and 
(he  parallel  sides  of  a  trapezoid  are  called  its  bases. 

LSI.  Def.  The  Altitude  of  a  parallelogram  or  trapezoid  is 
thr  perpendicular  distance  between  its  baa 

132.    Def.     The   Biaxial  of  a 
quadrilateral  La  a  straight  line  joining 

any  two  opposite  vertices. 


PROPOSITI-  \    WWII  I.       Thiohem. 

LS3.   /  ^a  par  all                         the  figure 

into  ftco  eq 

B  C 


A  I 

Let  ABC K  be  a  parallelogram,  and  A  C  its  diagonal. 
Wi  are  to  prove      A  A  B  C  =  A  A  E  <  . 
In  the  A  ABC  and  A  EC 

AC  =  AC,  Iden. 

Z  ACB  =  Z  CA  /:.  §68 

■inf.  A). 

ZCAB  =  ZACE,  §68 

.*.  AABC  =  AAEC,  §  107 

"  adj.  A  of  the  o  vely  to  a  wide  end  two 

adj.  A  of  the  other). 

Q.  E.  D. 


60  GEOMETRY. BOOK   I. 


Proposition  XXXIX.     Theorem. 

134.  In  a  parallelogram  the  opposite  sides  are  equal, 
and  the  opposite  angles  are  equal. 


A  E 

Let  the  figure  ABC E  be  a  parallelogram. 

We  are  to  prove     B  C  =  A  E,  and  AB  =  EC, 

also,     ZB  =  Z  E,andZ  BAE  =  Z  BCE. 

Draw  A  C 

A  ABC  =  AAEC,  §  133 

(the  diagonal  of  a  O  divides  the  figure  into  two  equal  A ). 

.'.BC  =  AE, 

and  AB  =  CE, 

(being  homologous  sides  of  equal  &  ). 

Z  B  =  Z  E, 

(being  Jiomologous  A  of  equal  A ). 

Z  BAC  =  Z  ACE, 

and  ZEAC  =  ZACB, 

(being  Jiomologous  A  of  equal  &)• 

Add  these  last  two  equalities,  and  we  have 

ZBAC  +  ZEAC  =  ZACE+ZACB; 

or,  ZBAE  =  ZBCE. 

Q.  E.  D. 

135.  Corollary.  Parallel  lines  comprehended  between  par- 
allel lines  are  equal. 


QUADRILATERALS.  61 


Proposition  XL.     Theorem. 

136.  If  u  quadrilateral  have  two  sides  equal  and  par- 
allel, then  the  otki  '  parallel,  and  the 
//// n  re  is  a  parallelogra m . 

B  C 


Let  the  figure  ABCE  be  a  quadrilateral,  having  the 
side  A  E  equal  and  parallel  to  BC. 

We  are  to  prove    A  B  equal  and  II  to  E  ( 

Draw  A  C. 

In  the  A  ABC  and  A  EC 

BC  =  A  A.  Hyp. 

A  C  =  AC,  [den. 

Z  BCA  =Z  CAE, 
{being alt. -int.  A). 

..AABC  =  AACE,  §  106 

'  sides  and  the  included  /Lof  tlie  <  to  two  sides 

and  the  included  Z.  of  the  M 

.\AB  =  EC, 
lologoits  sides  of  equal  A  ). 

Also,  ZBAC  =  ZACE, 

(being  homologous  A  of  equal  &  )  ; 

.'.A  Bis   II   to  EC,  §69 

/  lines  are  cut  by  a  third  straight  line,  if  the  alt.-int.  A  be 
equal  the  lines  are  pumUrl). 

.'.  the  figure  ABCEis&CJ,  §  125 

(the  opposite  sides  being  parallel). 

Q.  E    D. 


62  GEOMETRY. BOOK    I. 


Proposition  XLI.     Theorem. 

137.  If  in  a  quadrilateral  the  opposite  sides  be  equal,  the 
figure  is  a  parallelogram. 


A 

Let    the   figure    A  B  C  E    be   a    quadrilateral   having 
BC  =  AE  and  AB  =  EC. 

We  are  to  prove  figure     A  B  C  E  a  O. 

Draw  A  C. 

InthzA  ABC  and  A  EC 

BC  =  AE,  Hyp. 

AB  =  CE,  Hyp. 

A  C  =  A  C,  Iden. 

.'.  A  ABC  =  AAEC,  §  108 

(having  three  sides  of  the  one  equal  respectively  to  three  sides  of  the  other). 

.'./.  ACB  =  Z  CAEy 

and  ZBAC  =  ZACE, 

(being  homologous  A  of  equal  &  ). 

.'.BCis  II  toAE, 

and  A  Bis  II  to  EC,  §69 

(when  two  straight  lines  lying  in  the  same  plane  are  cut  by  a  third  straight 
line,  if  the  alt. -int.  A  be  equal,  tlie  lines  are  parallel). 

.'.  the  figure  ABC  E  is  a,  EJ,  §  125 

(having  its  opposite  sides  parallel). 

Q.  E.  D. 


QUADEILA1  BBAL8. 


63 


Proposition  XLII.     Thsobkm. 
138.   The  diagon  from  bisect  each  other. 

B  C 


Let   the  figure   A  B  C  E  be  a  parallelogram,  and  let 
the  diagonals  A  C  and  BE  cut  each  other  at  0. 

We  are  to  prove      AO  =  OC,  and  B  0=  0  ft 

In  the  A  A  OB  and  BOC 


AE=BC, 

§  134 

pposite  sides  of  a  E3\ 

ZOAE=ZOCB, 

§68 

(being  alt. -int.  A  ), 

Z  OEA=Z  OBC;  §  68 

.-.  AAOE  =  A  BOC,  §  107 

ng  a  side  and  two  adj.  A  of  the  <>  spectively  to  a  side  and  two 

adj.  A  of  the  other). 

..AO  =  OC, 

and  BO  =  OE. 

(being  homologous  sides  of  equal  &  ). 

Q.  E.  D. 


64  GEOMETRY. BOOK   I. 


Proposition  XLIII.     Theorem. 

139.   The  diagonals  of  a  rhombus  bisect  each  other  at 
right  angles. 

A  E 


Let    the  figure    A  B  C  E    he   a   rhombus,    having   the 
diagonals  A  C  and  BE  bisecting  each  other  at  0. 

We  are  to  prove      Z  AO E  and  Z  A  0  B  rt.  A. 

In  the  A  A  0  E  and  A  0  B, 

AE  =  AB,  §128 

(being  sides  of  a  rhombus)  ; 

OE=OB,  §138 

(the  diagonals  of  a  CD  bisect  each  otlier)  ; 

AO  =  AO,  Iden. 

.'.  A  AOE  =  A  AOB,  §  108 

(having  three  sides  of  the  one  equal  respectively  to  three  sides  of  the  other) ; 

.'.ZAOE  =  ZAOB, 

(being  homologous  A  of  equal  A  )  ; 
.\  Z  A  0  E  and  Z  A  0  B  are  rt.  A.  §  25 

When  one  straight  line  meets  another  straight  line  so  as  to  make  the  adj.  A 
equal,  each  Z  is  a  rt.  Z). 

Q.  E.  D. 


(»IAI>RILATERAI>.  65 


Proposition   XLIV.     Thboj 

1  10.    Two  paralltb  i   two   iifcs   and  the  in- 

cluded angle  ofth  %  and  the 

angle  of  the  other,  are  equal  in  all  resp* 

B  CD'  a 


A' 

In    the  parallelograms    A  B  C  D    and   A'  B1  O  D',    let 
AB  =  A' B',  AD  =  A'D',  and  Z.  A  =  Z  A'. 

We  are  to  prove  that  the  HJ  are  eq 

Apply  O  A  BCD  to  O  A'B'C'D',  so  that  A  D  will  fall 
<>ii  and  coinejdi  I  D'. 

Tli.:.  A  /I  will  fall  0] 
(/orZA  =  Z  A't  I 

and  B  will  fid!  on  B', 

B  =  A'  B\  \ 

Now,  BC  and  B' O  are  both  II  V  ad   an-  drawn 

through  point  B'\ 

.-.  tli-  line*  BCvbA  W  <"  coincide,  §  66 

ed 

In  like  manner  I>  0  and  />'  C  are  II  to  A'  11'  and  an  drawn 

through  tin-  point 

.-.  DC  and  /^  ' "  coincide :  §  66 

.-.  the  point  ('falls  on  I)'  C ,  oi  /> ' ' '•'  produced  ; 

.'.  (7  falls  on  both  AB'C; 

.*.  6'  must  fall  on  a  point  common  to  both,  namely,  C. 

.'.  the  two  UJ  coincide,  and  are  equal  in  all  respects. 

Q.  E.  D. 

Ml.   CoBOLLABT.    Two  red  vring  the  same  base  and 

aUitudi  <ir<>  ,-iu'it ;  fox  they  may  be  applied  to  each  other  and 
will  coincide. 


66  GEOMETRY. BOOK    I. 


Proposition  XLV.     Theorem. 

142.  The  straight  line  which  connects  the  middle  points 
of  the  non-parallel  sides  of  a  trapezoid  is  parallel  to  the  par- 
allel  sides,  and  is  equal  to  half  (heir  sum. 


C       U 

Let  SO  be  the  straight  line  joining  the  middle  points 
of  the  non-parallel  sides  of  the  trapezoid  ABCE. 

We  are  to  prove      SO  II  to  A  E  and  B  C ; 
also      SO  =  h(A  £  +  HC). 

Through  the  point  0  draw  F II  II  to  A  B, 

and  produce  B  C  to  meet  F  0 II  at  //. 

In  the  A  FOEbxA  C  0  II 

OE=OC,  Cons. 

ZOEF=ZOCII,  §68 

( fc  iiirj  alt.  -int.  A  ), 

ZFOE  =  ZCOH,  §49 

(being  vertical  A  ). 

/.A  FOE  =  A  CO//.  §  107 

{having  a  side  and  two  adj.  A  of  the  one  equal  respectively  to  a  side  and  two 
adj.  A  off/<<'  other). 


QUADRILATERALS.  67 


.\FM=CH% 

and  0F=01I, 

\g  homologous  sides  of  equal  A ). 

Now  F II  -  A  H  §  135 

(II  lines  conqrrelumded  between  II  lines  arc  equal)  ; 

..FO—AS.  Ax.  7. 

•\  the  figure  A  FO  Sis  a  O,  §  136 

(having  two  opposite  sides  equal 

.  .  .S'Ois  II  to  ^1  /■'.  §  125 

tig  opposite  sides  of  a  O). 

SO  is  also  II  to  BC, 

{a  st  i  I  to  one  of  two  II  lines  is  II  to  the  other  also). 


Now 

SO  =  A  I\ 
(being  opposite  sides  of  a  CJ)t 

§125 

and 

so  -  /;//. 

§  125 

But 

AF=AE-FE, 

and 

BH  =  BC+  C  II. 

Substitute  for  A  F  and  BE  their  equals,  ^i?  —  FE&ni 

BC+  Cll. 

and  add,  observing  that  C  //  =  F E\ 

then  2S0  =  AE+BC. 

.'.  SO  =  i(AE+  BC). 

Q.  E.  D. 


68 


GEOMETRY. BOOK  I. 


On  Polygons  in  General. 

143.  Def.  A  Polygon  is  a  plane  figure  bounded  by  straight 
lines. 

144.  Def.  The  bounding  lines  are  the  sides  of  the  polygon, 
and  their  sum,  s&AB  +  BC+CD,  etc.,  is  the  Perimeter  of 
the  polygon. 

The  angles  which  the  adjacent  sides  make  with  each  other 
are  the  angles  of  the  polygon. 

145.  Def.  A  Diagonal  of  a  polygon  is  a  line  joining  the 
vertices  of  two  angles  not  adjacent. 

B  B' 


C      A 


D      F' 


An  Equilateral  polygon  is  one  which  has  all  its 
An  Equiangular  polygon  is  one  which  has  all 


E 

146.  Def. 
sides  equal. 

147.  Def. 
its  angles  equal. 

148.  Def.  A  Convex  polygon  is  one  of  which  no  side, 
when  produced,  will  enter  the  surface  bounded  by  the  perimeter. 

149.  Def.  Each  angle  of  such  a  polygon  is  called  a  Salient 
angle,  and  is  less  than  two  right  angles. 

150.  Def.  A  Concave  polygon  is  one  of  which  two  or  more 
sides,  when  produced,  will  enter  the  surface  bounded  by  the 
perimeter. 

151.  Def.    The  angle  FD  E  is  called  a  Re-entrant  angle. 
AVhen  the  term  polygon  is  used,  a  convex  polygon  is  meant. 
The  number  offsides  of  a  polygon  is  evidently  equal  to  the 

number  of  its  angles. 

By  drawing  diagonals  from  any  vertex  of  a  polygon,  the  fig- 
ure may  be  divided  into  as  many  triangles  as  it  has  sides  less  two. 


POLYGONS.  69 


152.  I'ii.  Two  polygons  are  Equal,  when  they  can  be 
divided  by  diagonals  into  the  same  number  of  triangles,  equal 
each  to  each,  and  similarly  placed ;  for  the  polygons  can  be 
applied  to  each  other,  and  the  corresponding  triangles  will  evi- 
dently coincide.  Therefore  the  polygons  will  coincide,  and  be 
equal  in  all  respects. 

153.  Def.  Two  polygons  are  Mutually  1.  lar}  if  the 
angles  of  the  one  be  equal  to  the  angles  of  the  other,  each  to 
each,  when  taken  in  the  same  order;  as  the  polygons  ABODE F, 
and  A1  B1  C  D'  E'  F%  in  which  Z  A  -  Z  A1,  Z  B  =  Z  B', 
ZC  =  ZC,  etc. 

L5  1.  1  >i  i .  The  equal  angles  in  mutually  equiangular  poly- 
gons are  called  Homologous  angles;  and  the  sides  which  lie 
between  equal  angles  are  called  Homologous  sides. 

155.  Def.   Two  polygons  are  Mute  //,  if  the 

sides  of  tie  tl  to  the  sides  of  the  other,  each  to  each, 

when  taken  in  the  same  order. 


Fig.   1.  Pig.  3.  :.  4. 

Two  polygons  may  be  mutually  equiangular  without  being 
mutually  equflatei  I  and  2. 

And,  t*f  two  polygons  may  be 

mutually    equilateral    without  being   mutually   equiangular \  as 
S  and  4. 

If  two  polygons  be  mutually  equilateral  and  equiangular, 
they  are  equal,  for  they  may  be  applied  the  one  to  the  othei 
as  to  coincide. 

156.  Def.    A  polygon  of  three  sides  is  a  Trigon  or  Tri- 
angle ;  one  of  f«»ur  sides  is  a  Tetragon  or  Quadrilateral ;  one  of 
fagon ;  one  of  six  sides  is  a  Hexagon ;  one  of 

ii  sides  is  a  Heptagon;  one  of  eight  sides  is  an  Octagon;  one 
of  fcen  agon  ;  one  of  twelve  sides  is  a  Dodecagon. 


70 


GEOMETRY. BOOK  I. 


Proposition  XLYI.     Theorem. 

157.  The  sum  of  the  interior  angles  of  a  polygon  is 
equal  to  two  right  angles,  taken  as  many  times  less  two  as 
the  figure  has  sides. 


A  B 

Let  the  figure  ABC DEF  be  a  polygon  having  n  sides. 

We  are  to  prove 

A  A  +  Z  B  +  A  C,  etc.,  =  2  rt.  A  (n  -  2). 
From  the  vertex  A  draw  the  diagonals  A  C,  A  D,  and  A  E. 

The  sum  of  the  A  of  the  A  =  the  sum  of  the  angles  of  the 
polygon. 

Now  there  arc  («  —  2)  A, 


and  the  sum  of  the  A  of  each  A  =  2  rt.  A. 


§98 


.'.  the  sum  of  the  A  of  the  A,  that  is,  the  sum  <>f  the  A  of 
the  polygon  =  2  rt.  A  (n  —  2). 


Q.  E.  D. 


158.  Corollary.    The  sum  of  the  angles  of  a  quadrilateral 

equals  two  right  angles  taken  (4  —  2)  times,  i.  e.  equals  4  right 
angles;  and  if  the  angles  be  all  equal,  each  angle  is  a  right 
angle.     In  general,  each  angle  of  an  equiangular  polygon  of  n 

9  (v  —  9\ 
is  equal  to  ^-A_ J.  right  angles. 


POLYGONS.  71 


PBOPOemOH  XLVIL     Theorem. 

159.  Tl  i  of  a  polygon,  nufde  by  produ- 

cing each  <  if  lis  -svVA'.y  in  succession,  arc  together  equal  to  four 
rig/tt  angles. 


Let  the  figure  ABODE  be  a  polygon,  having  its  sides 
produced  in  succession. 

•IPJ   //    to  prove  the  sum  oj  A. 

Denote  the  int.  A  of  the  polygon  by  A,B,t\  i>.  E  \ 

ami  the  ext  .    1  ■.  </,  e. 

Z  A  +  Za  =  2vtA, 

.  <4 ). 

Z  B  +  Z  h  =  2  rt.  A  $  84 

In  like  manner  each  pair  ofacjj.  A  =  2  rt.  zS; 

■\  tin'  sum  <»f  the  interior  and  exterior  .  i  •»  2  rt  -d  taken 

as  many  t ;: 

it.  A. 
the   interior  .  2  rt.  .  S  taken   as   many  times  as  the 

figure  has  aides  Less  two,       2  rt  A  («.  —  2), 

it.  .j  —  1  rt  A. 

.■ .  the  exterior  z^  =  -4  rt  A. 

Q.  E.  D. 


72 


GEOMETRY. BOOK    I. 


Exercises. 

1.  Show  that  the  sum  of  the  interior  angles  of  a  hexagon  is 
equal  to  eight  right  angles. 

2.  Show  that  each  angle  of  an  equiangular  pentagon  is  £  of 
a  right  angle. 

3.  How  many  sides  has  an  equiangular  polygon,  four  of 
whose  angles  are  together  equal  to  seven  right  angles'? 

4.  How  many  sides  has  the  polygon  the  sum  of  whose  in- 
terior angles  is  equal  to  the  sum  of  its  exterior  angles  1  ^t    Q/V 

■  .  How  many  sides  has  the  polygon  the  sum  of  whose  in- 
terior angles  is  double  that  of  its  exterior  angles?       U      OiA^-J % 

6.  How    many   sides   has   the   polygon   the   sum   of  whose, 
rior  angles  is  double  that  of  its  interior  angles'?    ^&t-     O^J { 

7.  Every  point  in  the  bisector  of  an  angle  is  equally  distant 
from  the  sides  of  the  angle  ;  and  every  point  not  in  the  bisector, 
but  within  the  angle,  is  unequally  distant  from  the  sides  of  the 
angle. 

8.  B  A  C  is  a  triangle  having  the  angle  B  double  the  angle 
A.  If  B  D  bisect  the  angle  B,  and  meet  A  C  in  D,  show  that 
B  D  is  equal  to  A  D. 

9.  If  a  straight  line  drawn  parallel  to  the  base  of  a  triangle 

the  sides,  show  that  it  bisects  the  other  also  ;  and 
that  the  portion  of  it  intercepted  between  the  two  sides  is  equal 
to  one  half  the  base. 

\  10.    A  B  C  J)  is  a  parallelogram,  A' and  F  the  middle  points 
of  AD  and   BC  respectively;  show  that  BE  and   DF  will 
ct  the  diagonal  A  C. 

11.  If  from  any  point  in  the  base  of  an  isosceles  triangle 
parallels  to  the  equal  sides  he  drawn,  show  that  a  parallelogram 
is  formed  whose  perimeter  is  equal  to  the  sum  <>("  the  equal 

s  of  the  triangle. 

12.  If  from  the  diagonal  11 1)  of  a  square  A  BCD,  BE  be 
•  >fT  equal  to  BC,  and  E F  }>•■  drawn  perpendicular  to  BDy 

show  that  D  E  is  equal  to  F  F,  and  also  to  FC. 

1  3.  Show  that  the  three  lines  drawn  from  the  vert  ices  of  a 
triangle  to  the  middle  points  of  the  opposite  sides  meet  in  a 
point 


BOOK  II. 

CIRCLES. 


PiaiM'I  , 

160.   I  >  i :  i .    A  '  a  plane  figure  bounded  by  a  curved 

line,  all  the  points  of  which  are  equally  distanl  from  a  point 
within  called  ill'*  & 

1''.  1.  I  > i  i .  1  nee  of  a  circle  is  the  line  which 
bounds  tin-  qu 

162.  of  a  circle  is  any  straight  line  drawn 
tram  the  centric  t o  the  circumference,  as  0  -1.  1 

163.  1 1  (meter  of  a  cii<  line  paw- 

be  centre  and  havh  ircum- 

. 
By  :li    definition  of  a  circle,  all  its  radii  are  equal.     He  i 
all  i;  .  since  tin-  dian  I  to  twice 

i  liua 


M  .1/ 


16  I.    1  »i  i .    An  Arc  of  a  circle  is  any  portion  of  the  circum- 

is  an  arc  equal  to  one 
half  the  circumference,  as  A  M  l>.  Fig.  2. 

166.    Dl  >rd  of  a  circle  is  any  straight  line  having 

:ivniiti<s  in  the  circumference,  as  A  },. 

iy  chord  subtends  two  arcs  whose  sum  is  the  cir- 
cumference. Thus  the  chord  .1  A'.  (Fig.  3),  subtends  the  arc 
.!  M/>  ami  the  aw  A  I)  11.  Whenever  a  chord  and  its  arc  are 
ipoken  of,  the  I  Is  meant  unless  it  be  otherwise  stated. 


74  GEOMETRY. BOOK   II. 

167.  Def.  A  Segment  of  a  circle  is  a  portion  of  a  circle 
enclosed  by  an  arc  and  its  chord,  as  A  M  B,  Fig.  1. 

168.  Def.  A  Semicircle  is  a  segment  equal  to  one  half  the 
circle,  as  A  B  C,  Fig.  1. 

169.  Def.  A  Sector  of  a  circle  is  a  portion  of  the  circle 
enclosed  by  two  radii  and  the  arc  which  they  intercept,  as  A  C  B, 
Fig.  2. 

170.  Def.  A  Tangent  is  a  straight  line  which  touches  the 
circumference  but  does  not  intersect  it,  however  far  produced. 
The  point  in  which  the  tangent  touches  the  circumference  is 
called  the  Point  of  Contact,  or  Point  of  Tangency. 

171.  Def.  Two  Circumferences  are  tangent  to  each  other 
when  they  are  tangent  to  a  straight  line  at  the  same  point. 

172.  Def.  A  Secant  is  a  straight  line  which  intersects  the 
circumference  in  two  points,  as  A  /),  Fig.  3. 


173.  Def.    A  straight  line  is  Inscribed  in  a  circle  when  its 
extremities  lie  in  the  circumference  of  the  circle,  as  A  B,  Fig.  1. 

An  angle  is  inscribed  in  a  circle  when  its  vertex  is  in  the 
circumference  and  its  sides  are  chords  of  that  circumference,  as 
Z  ABC,  Fig.  1. 

A  polygon  is  inscribed  in  a  circle  when  its  sides  are  chords 
of  the  circle,  as  A  A  B  C,  Fig.  1. 

A  circle  is  inscribed  in  a  polygon  when  the  circumference 
touches  the  sides  of  the  polygon  but  does  not  intersect  them, 
as  in  Fig.  4. 

174.  Def.    A  polygon  is  Circumscribed  about  a  circle  when 
all  the  sides  of  the  polygon  are  tangents  to  the  circle,  as  in  Rg,  4. 

A  circle  is  circumscribed  about  a  polygon  when  the  circumfer- 
ence .passes  through  all  the  vertices  of  the  polygon,  as  in  Fig.  1. 


STRAIGHT    LINES    AND    CIRCLES. 


75 


175.  Def.    Equal  circles  are  circles  which  have  equal  radii. 
For  if  one  circle  be  applied  to  the  other  so  that  their  centres 
coincide  their  circumferences  will  coincide,  since  all  the  points 
both  aiv  at  the  same  distance  from  the  centre. 

1  7*  >.  Every  diameter  bisects  the  circle 
a n< I  its  circumference.  For  if  we  fold  over 
I  MB  on  A  B  as  an  axis  until 
it  comes  into  the  plane  of  A  P  B,  the  arc 
A  MB  will  coincide  with  the  arc  APB; 
because  every  point  in  each  is  equally  dis- 

r  i *  'in  the  '•nitre  0. 


PBOFOflmoi    1.     Theorem. 


17  7.   The  diameter  of  a  circle  i$  greater  than  n,u/  other 

clnnil . 

Let  A  H  be  the  diameter  of  the  circle 
A  MB,  and  A  E  any  other  chord. 

II      '      to  prove     A  B  >  A  /'. 

From  C,  the  centre  of  the  O,  draw  C  E. 
CE=CB, 
(being  radii  of  the  same  circle). 

But  AC+CE>AEy  §96 

(the  mm  of  two  sides  of  a  A>  the  third  side). 

Substitute  for  C  E,  in  the  above  inequality,  its  equal  CB. 

Then  AC+CB>AJB,ot 

A  B  >  AE. 

Q.  E.  D. 


76 


GEOMETRY. BOOK   II. 


Proposition  II.     Theorem. 

178.  A  straight  line  cannot  intersect  the  circumference 
of  a  circle  in  more  than  two  points. 


Let  II K  be  any  line  cutting  the  circumference  A  MP. 

We  are  to  prove  thai  UK  can   intersect  the  circinnference 
in  only  two  points. 

If  it  he  possible,  let  II K  intersect  the  circumference  in  three 
points,  //,  P,  and  K. 

From  0,  the  centre  of  the  O,  draw  the  radii  OH,   OP, 
and  0  K. 


Then  0 II,  0  P,  and  0  K  are  equal, 

(being  radii  of  the  same  circle). 


§163 


.*.  if  UK  could  intersect  the  circu inference  in  three  points, 
we  should  have  three  equal  straight  lines  01/,  OP,  and  OK 
drawn  from  the  same  point  to  a  given  straight  line,  which  is 
impossible,  §  56 

(only  two  equal  straight  lilies  can  be  drawn  from  a  point  to  a  straight  line), 

.'.a  straight  line  can  intersect  the  circumference  in  only 
two  points. 

Q.  E.  D. 


STRAIGHT    LINES    AND    CIRCLES.  77 


^ 


Proposition  III.     Theorem. 


179.    In   tin  rcle,  or  equal  .  equal  angles 

at  the  centre  intercept  equai  arcs  an  the  circmnfference. 


P  F 

In  the  equal  circles  ABP  and  A'B'P'  let  ZO=ZO'. 

We  are  to  prove      arc  RS  =  arc  R'  S1. 

Apply  O  A  BP  to  O  A'B'P, 

so  that  Z  0  shall  coincide  with  Z.  0'. 

The  point  R  will  fall  upon  R'9  §  176 

(for  0R=  0'  Rf,  being  radii  of  equal  <D), 

1  the  point  £  will  fall  upon  S',  §  176 

(for  0S=  0'&,  being  radii  of  eqyal  ©). 

Then  the  arc  R  S  must  coincide  with  the  arc  R'S'. 
For,  otherwise,  there  would  be  some  points  in  the  circumference 
unequally    distant  from  the  centre,  which   is  contrary  to   the 
definition  of  a  circle.  §  160 

Q.  E.  D. 


78  GEOMETRY. BOOK   II. 


Proposition  IV.     Theorem. 

180.  Conversely  :  In  the  same  circle,  or  equal  circles } 
equal  arcs  subtend  equal  angles  at  the  centre. 


In  the  equal   circles  ABP    and  A' B' P'   let   arc    RS 
=  arc  R'S1. 

Wearetoprove     Z  ROS  =  Z  R' 0' S'. 

Apply  Q  ABP  to  O  A' B' /», 

so  that  the  radius  0  R  shall  fall  upon  0'  R'. 

Then  S,  the  extremity  of  arc  RS, 

will  fall  upon  S',  the  extremity  of  arc  R'  S', 
(for  BS=B>S',  by  hyp.). 

.'.OS  will  coincide  with  0'  S',  §  18 

(their  extremities  being  the  saine  points). 

.'.  Z  R  0  S  will  coincide  with,  and  be  equal  to,  Z  R'  0'  S'. 

Q.  E.  D. 


STRAIGHT    LINKS    AND    CIRCLES. 


79 


Proposition  V.     Theorem. 

181.  In  the  same  circle,  or  equal  circles,  equal  arcs  are 
subtended  by  equal  chords. 


In    the    equal   circles  ABP   and  A'  B'  V  let   arc   RS 
=  arc  R'S'. 

We  are  to  prove      chord  R  S  =»  chord  R'  S'. 

Draw  the  radii  0  R,  0  S,  0'  R'y  and  0'  S'. 

In  the  A  R  0  S  and  R'  0'  & 


OR=( 
(being  radii  of  equal  < 

OS=0'S', 


§176 

§176 
§  180 


ZO  =  ZO', 

0  subtend  equal  A  at  the  centre). 

.\AROS  =  A  R'O'S',  §  106 

{two  sides  and  the  included  Z  of  the  one  being  equal  respectively  to  two  sides 
and  the  included  Z.  of  the  other). 

.'.  chord  RS=  chord  R'S1, 
(being  homologous  sides  of  equal  &  ). 

Q.  E.  D. 


8U 


GEOMETRY. BOOK    II. 


Proposition  VI.     Theorem. 

182.  Conversely  :    In  ike  same  circle,  or  equal  circles^ 
equal  chords  subtend  equal  arcs. 


In  the  equal  circles  ABP  and  A' B' P',  let    chord  RS 
=  chord  R'S'. 

We  are  to  prove      arc  R  S  =  arc  R'  S'. 

Draw  the  radii  0  R,  0  S,  0'  R',  and  0'  S'. 

In  the  A  R  0  S  and  R' 0' S' 

RS  =  R'S', 


on^O'R', 

{being  radii  of  equal  (D), 

OSLO'S' ; 


Hyp. 
§  176 

§176 
§108 


.-.  A  ROS  =  A  R'O'S', 
(three  sides  of  the  one  being  equal  to  three  sides  of  the  otlicr). 

:.£  0  =  Z  0', 

(being  homologous  A  of  equal  4). 

.'.arc  RS  =  arc  IPS',  §  179 

(in  the  same  O,  or  equal  (D,  eqical  A  at  the  centre  intercept  equal  arcs  on  the 

circumference), 

Q.  E.  D. 


STRAIGHT    LINKS    AND    CIRCLES. 


81 


ItTf 


PROPOSITION    ^S   II. 


Theorkm. 


183.  The  radius  perpendicular  to  a  c/wrd  bisects  tie 
chord  and  the  arc  /  it. 


Lft 


Let  A  B  be  the  chord,  and  let  the  radius  C S  be  per- 
pendicular to  A  B  at  the  point  M. 

We  are  to  prove      A  M '=  B M,  and  arc  A  S  =  arc  B  S. 

Draw  CA  and  C  /!. 

CA  =  CB, 

(being  radii  of  the  same  O) ; 

.*.  A  AC  B  is  isosceL  §  84 

(the  opposite  sides  being  equal)  ; 

.'.  _L  CS  bis  A  B  and  the  Z  (7,         §  113 

(the  JL  drawn  from  the  vertex  to  the  base  of  an  isosceles  A  bisects  the  base  and 
the  Z.  at  the  vertex). 

.\AM=  BM. 

Also,  since  ZACS  =  ZBCS, 

arc  A  S  =  sac  SB,  §179 

(equal  A  at  the  centre  intercept  equal  arcs  on  the  circumference). 

Q.  E.  D. 

184.  Corollary.  The  perpendicular  erected  at  the  middle 
of  a  chord  passes  through  the  centre  of  the  circle,  and  bisects 
the  arc  of  the  chord. 


82  GEOMETRY. BOOK   II. 


Proposition  VIII.     Theorem. 

185.  In  the  same  circle,  or  equal  circles,  equal  chords 
are  equally  distant  from  the  centre ;  and  of  two  unequal 
chords  the  less  is  at  the  greater  distance  from  the  centre. 


In  the  circle  A  B EC  let  the  chord  A  B  equal  the  chord 
C F,  and  the  chord  C  E  be  less  than  the  chord  G F. 
Let  OP,  OH,  and  0 K  be  J*  drawn  to  these  chords 
from  the  centre  0. 

We  are  to  prove     OP=OH,  and  OH<  OK. 

Join  OA  and  OC. 
In  the  rt.  A  A  0 P  and  CO  H 

OA  =  OC, 

(being  radii  of  tlie  same  O)  ; 

AP=CH,  §183 

(being  halves  of  equal  chords) ; 

.\AAOP  =  A  COI/,  §  109 

yivo  rt.  &  are  equal  if  they  have  a  side  and  hypotenuse  of  the  one  equal  to 
a  side  and  hypotenuse  of  the  other). 

.'.  OP=OII, 

(being  homologous  sides  of  equal  &). 
Again,  since  C  E  <  C  F, 

the  J_  0  K  will  intersect  C  F  in  some  point,  as  m. 
Now  OK>Om.  Ax.  8 

But  Om>OII,  §52 

(a  A.  is  the  shortest  distance  from  a  point  to  a  straight  line). 

.'.  much  more  is  0  K>  Oil. 

Q.  E.  D. 


STRAIGHT    LINES    AND    CIRCLES. 


83 


Proposition  IX.     Theorem. 

186.  A  straight  line  perpendicular  to  a  radius  at  its 
nifij  is  a  tangent  to  the  circle. 


Let  BA    be   the   radius,    and  MO    the   straight  line 
perpendicular  to  BA  at  A. 

We  are  to  prove      M  0  tangent  to  the  circle. 

From  B  draw  any  other  line  to  M  0,  as  B  C  H. 

IlIf>BA,  §52 

(a  ±  measures  the  shortest  distance  from  a  point  to  a  straight  line). 

.*.  point  H  is  without  the  circumference. 

But  B  II  ifl   my  other  line  than  B  A, 

.*.  every  point  of  the  line  MO  is  without  the  circumference, 
except  A. 

.*.  MO  is  a  tangent  to  the  circle  at  A.  §  171 

Q.  E.  D. 

187.  Corollary.  When  a  straight  line  is  tangent  to  a 
circle,  it  is  perpendicular  to  the  radius  drawn  to  the  point  of 
contact,  and  therefore  a  perpendicular  to  a  tangent  at  the  point 
of  contact  passes  through  the  centre  of  the  circle. 


84 


GEOMETRY. BOOK   II. 


Proposition  X.     Theorem. 

188.  When  two  circumferences  intersect  each  other,  the 
line  which  joins  their  centres  is  perpendicular  to  their  common 
chord  at  its  middle  point. 


Let  C  and  C  be  the  centres  of  two  circumferences 
which  intersect  at  A  and  B.  Let  A  B  be  their 
common  chord,  and  C  C  join  their  centres. 

We  are  to  prove  C  C  A.  to  A  B  at  its  middle  point. 

A  J_  drawn  through  the  middle  of  the  chord  A  B  passes 
through  the  centres  C  and  C,  §  184 

(a  ±  erected  at  the  middle  of  a  chord  passes  through  the  centre  of  the  O). 
•\  the  line  C  C,  having  two  points  in  common  with  this  J_, 
must  coincide  with  it. 


.".  C  C  is  _L  to  A  B  at  its  middle  point. 


Q.  E.  D. 


Ex.  1.  Show  that,  of  all  straight  lines  drawn  from  a  point 
without  a  circle  to  the  circumference,  the  least  is  that  which, 
when  produced,  passes  through  the  centre. 

Ex.  2.  Show  that,  of  all  straight  lines  drawn  from  a  point 
within  or  without  a  circle  to  the  circumference,  the  greatest  is 
that  which  meets  the  circumference  after  passing  through  the 
centre. 


STRAIGHT    I.IN'ES    AND   CIF 


$0 


UNIVERSITY 

Proposition  XI.     Theorb^C  J  ^TPftTL^^ 

189.   //7/r//  ftp*  cirewntfen  i '  mr/entTo  $&dk  other 

theif  point   of  contact   M  in    the  straight   line  joining    their 
rp  nt  rex. 


Let  the  two  circumferences,  whose  centres  are  C  and 
C,  touch  each  other  at  0,  in  the  straight  line  A  B, 
and  let  CC  be  the  straight  line  joining  their  cen- 
tres. 

We  are  to  prove      0  is  in  tlie  straight  luu  0  C. 

A  _L  to  .1  11.  «lia\\n  through  the  point  0,  passes  through  the 
centres  £7  and  C,  §  187 

{a  _L  to  a  tangent  at  the  point  of  contact  passes  through  tlie  centre  of  the  O). 

.'.  the  line  C  C>  halving  two  points  in  common  with  this  _L, 
must  coincide  with  it. 


.'.  0  is  in  Am  straight  Line  C  C. 


Q.  E.  D. 


F,\\   A  />,    a   chord   of  a   circle,   is  the  hase  of  an   isosceles 

triangle  whose  vertex  0  is  without  the  circle,  and  whose  equal 

meet   the  circle  in   1>  and   E.     Show  that  CD  is  equal 

to  c  /■/. 


86  GEOMETRY. BOOK   II. 


On  Measurement. 

190.  Def.  To  measure  a  quantity  of  any  kind  is  to  find 
how  many  times  it  contains  another  known  quantity  of  the  same 
hind.  Thus,  to  measure  a  line  is  to  find  how  many  times  it  con- 
tains another  known  line,  called  the  linear  unit. 

191.  Def.  The  number  which  expresses  how  many  times 
a  quantity  contains  the  unit,  prefixed  to  the  name  of  the  unit, 
is  called  the  numerical  measure  of  that  quantity ;  as  5  yards,  etc. 

192.  Def.  Two  quantities  are  commensurable  if  there  be 
some  third  quantity  of  the  same  kind  which  is  contained  an 
exact  number  of  times  in  each.  This  third  quantity  is  called 
the  common  measure  of  these  quantities,  and  each  of  the  given 
quantities  is  called  a  multiple  of  this  common  measure. 

193.  Def.  Two  quantities  are  incommensurable  if  they 
have  no  common  measure. 

194.  Def.  The  magnitude  of  a  quantity  is  always  relative 
to  the  magnitude  of  another  quantity  of  the  same  kind.  No 
quantity  La  great  oi  small  except  by  comparison.     This  relative 

aitude  is  called  their  Ratio,  and  this  ratio  is  always  an  ab- 
stract number. 

When  two  quantities  of  the  same  kind  are  measured  by  the 
same  unit,  their  ratio  is  the  ratio  of  their  numerical  measures. 

195.  The  ratio  of  a  to  6  is  written  -,  or  a  :b,  and  by  this 

mt  : 

How  many  times  b  is  contained  in  a;  a 

mi.  what  part  a  is  of  b.  b 

I.  If  b  be  contained  an  exact  number  of  times  in  a  their 
ratio  is  a  whole  number. 

If  b  be  not  contained  an  exact  number  of  times  in  a,  but 
if  there  be  a  common  measure  which  is  contained  m  times  in  a 

m 

and  ii  times  in  b,  their  ratio  is  the  fraction  — . 

n 

II.  If  a  and  b  be  incommensurable,  their  ratio  cannot  be 
ictly  expressed  in  figures.     But  if  b  be  divided  into  n  equal 

parts,  and  one  of  these  parts  be  contained  m  times  in  a  with 

i   remainder  less  than  -  part  of  b,  then  —  is  an  approximate 
n  n 

.a  1 

value  of  the  ratio  -,  correct  within 


THBOBY    of    LIMITS.  87 

Again,  if  each  of  these  equal  parts  of  b  be  divided  into  n 

equal  parts;  that  is,  if  b  be  divided  into  n2  equal  parts,  and  if 

of  these  parts  be  contained  m'  times  in  a  with  a  remainder 

1  m! 

less  than  -j  part  of  b,  then  — ^  is  a  nearer  approximate  value 

of  the  ratio  -,  correct  within  -s-. 
6  n 

continuing  this   process,  a  series  of  variable  values, 

Ttl         111}       Til/* 

—  >—?>—*>  etc.,  will  be  obtained,  which  will  differ  less  and 
n       n2      n*  a 

from  the  exact  value  of  -.     We  may  thus  find  a  fraction 

which  shall  differ  from  tin-  due  by  as  little  as  we  please, 

that  is,  bj  leei  than  any  m  uantity. 

ICO,  an   incommenstn  i   is  the  limit  toward  which 

its  successive  approximate  values  are  constantly  tending. 

On   im    Theory  of  Limits. 

l»       When  tttil        regarded  as  haying  a  fixed 

value,  it  is  called  a  c  but,  when  it  is  regarded,  under 

the  oonditions  imposed  upon  it,  as  bavin  nxber 

fereni  values,  it  is  called  a  I 

197.  Dl  i.  When  it  ran  he  shown  that  the  value  of  a  vari- 
able, i  at  a  series  of  intervals,  can  by  indefinite 

nuation  of  the  series  be  made  to  differ  from  a  given  con- 
stant by  less  than  any  assigned  quantity,  h  mall,  hut 
cannot  be  made  absolutely  equal  to  the  constant,  that  constant 

lied  the  Limit  of  ible,  and  the  variable  is  said  to 

If  the  variable  be  increasing,  its  limit  is  called  a  superior 
limit;  if  decreasing,  tor  limit 

198.  Suppose  a  point  ± * *      *'    B 

to  move  tiom  A  toward  B,  umlt-r  the  conditions  that  the  first  sec- 
ond it  shall  move  one-half  the  distance  from  A  to  B,  that  is, 
to  M\  the  next  second,  one-half  the  remaining  distance,  that  is, 
to  M' ;  the  next  second,  one-half  the  remaining  distance,  that 
is,  to  M"s  and  so  on  indefinitely. 

Then  it  is  evident  that  the  moving  point  may  approach  as 
but  will  never  arrive  at  B.     For,  however 


88  GEOMETRY. BOOK    II. 

near  it  may  be  to  B  at  any  instant,  the  next  second  it  will  pass 
over  one-half  the  interval  still  remaining  ;  it  must,  therefore, 
approach  nearer  to  B,  since  half  the  interval  still  remaining  is 
some  distance,  but  will  not  reach  B,  since  half  the  interval  still 
remaining  is  not  the  whole  distance. 

Hence,  the  distance  from  A  to  the  moving  point  is  an  in- 
creasing variable,  which  indefinitely  approaches  the  constant  A  B 
as  its  limit ;  and  the  distance  from  the  moving  point  to  B  is  a 
decreasing  variable,  which  indefinitely  approaches  the  constant 
zero  as  its  limit. 

If  the  length  of  A  B  be  two  inches,  and  the  variable  be 
denoted  by  x,  and  the  difference  between  the  variable  and  its 
limit,  by  v  : 

after  one  second,         x  =  1 ,  v  =  1  : 

after  two  seconds,       a?  =  1  H-  J,  v==i  ] 

r  three  seconds,     a?  =  1  +  £  +  J,  v  =  \\ 

after  four  seconds,       #=1 +  £  +  £  +  £,    v  =  \\ 

and  so  on  indefinitely. 
Now  tha  sum  of  the  series  1  +  £  +  \  +  \  etc.,  is  evidently 
less  than  2;  but  by  taking  a  great  number  of  terms,  the  sum 
can  be  made  to  differ  from  2  by  as  little  as  we  please,  Hence 
2  is  the  limit  of  the  sum  of  the  Beries,  when  the  number  of  the 
terms  is  increased  Indefinitely  ;  and  0  is  the  limit  of  the  variable 
difference  between  this  variable  sum  and  2. 

Urn.  will  be  used  as  an  abbreviation  for  limit. 

199.  [1]  The  difference  between  a  variable  and  its  limit  is  a 
variable  whose  limit  is  zero. 

[2]  If  two  or  more  variables,  v,  v*,  v",  etc.,  have  zero  for  a 
limit,  their  sum,  v+  v1  +  v",  etc.,  will  have  zero  for  a  limit. 

[3]  If  the  limit  of  a  variable,  v,  be  zero,  the  limit  of  a±v 
vill  be  the  constant  a,  and  the  limit  of  a  X  v  will  be  zero. 

[4]  The  product  of  a  constant  and  a  variable  is  also  a  va- 
riable,  and  the  limit  of  the  product  of  a  constant  and  a  variable 
is  the  product  of  the  constant  and  tht  limit  of  the  variable. 

[5]  The  sum  or  product  of  two  variables,  both  of  which  am 
i  ither  <  ilso  a  variable. 


THEORY    OF    LIMITS.  89 


Proposition  I. 
[G]  If  two  variables  be  always  equal,  their  limits  are  equal. 

Let  the  two  variables  A  M  and 
A  N  be  always  equal,  and  let  A  C 
and  A  It  be  their  respective  limits. 

We  are  to  prove  A  C '  =  A  B. 

Suppose  AOAB.  Then  we  may 
diminish  A  C  to  boom  value  A  C  Bach 
that  A  <'  =  AB. 

Since  A  M  approaches  indefinitely  to 
A  C,  we  may  suppose  thai  it  ihed 

■  value  A  P  greater  than  A  C. 

.\  Q  be  the  corresponding  value  of  A  N. 

Then  A  /'      A  Q. 

A  G"  -  A  ft 
both  of  these  equations  can  rue,  for  A  P>  A  C, 

and  /l  (>  <  J  ft     .*.  A  C  cannot  be  greater  than  A  B. 

Ag*ain,  suppose  A  C<  A  B.  Then  we  may  diminish  A  B  to 
some  value  A  B'  such  that  A  0m ■   i 

since  il  iV  approaches  indefinil  B  we  may  suppose 

that  it  has  reached  a  value  A  Q  greater  than  A 

Let  AP  be  t!  ponding  value  of  J  Jbf. 

Then  A  r  -  A  <>. 

AC=  A  A. 

Bat  both  of  the©  ■  lie,  for  .1  P<  J  C, 

and  J  Q  >  .1  B .     •'.  .1  C  cannot  be  leea  than  A  ft 

sin  !iiu»t  be  greater  or  lees  than  A  JJ,  it  must  be 

equal  to  .1  ft  QED- 

[7]  Corollary    1.    If  two  variables  be  in  a  constant  rutin, 
■he  same  ratio.     Fi  ft,  let  x  and  y  be  two  variables 

x 

having  the  constant  ratio  r,  then  -  =  r,  or,  x  =  r  y,  therefore 

y 

)  =  Urn.  (ru)  =  rX  Urn.  (y).  therefore  ,.    '  ,   t  =  r. 
v   ;  v    y/  vy/'  /tm.  (y) 

[8]  Cor.  2.  Since  an  incommensurable  ratio  is  the  limit  of 
its  successive  approximate  values,  two  incommensurable  ratios  - 
and        '//<  '  I'i'il  ij  t/tey  always  have  the  same  approximate  values 

>xjyressed  within  tin  satiu  measure  of  precisian. 


90  GEOMETRY. BOOK  II. 


Proposition  II. 

[9]  The  limit  of  the  algebraic  sum  of  tivo  or  more  variables 
is  the  algebraic  sum  of  their  limits. 

Let  x,  y,  z,  be  variables,  a,  b,  and  c,    a +-- 

their  respective  limits,  and  v,  v',  and  v", 

the  variable  differences  between  x.  y.  z,    b +— 

and  a,  by  c,  respectively. 

We  are  to  prove  Urn.  (x  +  y  +  z)  =  a  +  b  +  c.     c *•— - 

Now,  x  =  a  —  v,  y  —  b  —  v*,  zz=c  —  v". 
Then,  x  -f-  y  +  z  =  a  —  v+  b  —  v*  +  c  —  ?/'. 

.*.  lim.{x  +  y  +  z)=lim.(a — v+b — vIJrc—  v").    .  [G] 

But,    Urn.  {a-v  +  b  —  v'  +  c  —  v")  =  a+b  +  c.  [3] 

.*.  Urn.  (x  H-  y  +  z)  =  a  -h  b  H-  c. 

Q.  E.  D. 

Proposition  III. 

[10]  The  limit  of  the  product  of  two  or  more  variables  is  the 
product  of  their  lint  its. 

Let  x,  y,  z,  be  variables,  a,  6,  c,  their  respective 
limits,  and  v,  i/,  v",  the  variable  differences  between 
x,  y,  z,  and  a>  b,  c,  respectively. 

We  are  to  prove  Urn.  (xy  z)=1  a  b  c. 
Now,  #  =  a  —  v,  y  =  b  —  v',  z  =  c  —  v". 
Multiply  these  equations  together. 

Then,  xyz  =  abc^f  terms  which  contain  one  or  more  of 
the  factors  v,  V,  v",  and  hence  have  zero  for  a  limit.  [31 

.*.  Urn.  (xyz)  =  Urn.  (a  bc^f  terms  whose  limits  are  zero).  [GJ 
But  lim.  (ab  c  =F  terms  whose  limits  are  zero)  —  a  b  c. 
. ' .  Urn.  (xyz)  =  a  b  c. 

Q.  E.  D. 

For  decreasing  variables  the  proofs  are  similar. 


Note.  —  In  the  application  of  the  principles  of  Limits,  refer- 
ence to  this  section  (§  199)  will  always  include  the  fundamental 
truth  of  limits  contained  in  Proposition  I.  ;  and  it  will  be  left  as 
an  exercise  for  the  student  to  determine  in  each  case  what  other 

truths  of  this  section,  if  any,  are  included  in  the  reference. 


MEASUREMENT    OF    ANGLES. 


91 


Proposition  XII.     Theorem. 

200.  In  the  same  circle,  or  equal  circlet,  two  commen- 
turable  arcs   have  the  same  rutin  as  the  angles   which  they 
in!  at  the  centre. 


H. 


.K 


In  the  circle  A  PC  let  the   two  arcs  be  A  B  and  A  C, 
and  A  0  B  and  A  OC  the  A  which  they  subtend. 

Z  A  OB 


We  are  to  prove 


a  /; 


arc  AC       ZTiOQ 


Let  U  K  be  a  common  measure  of  A  B  and  A  C. 

Suppose  //  K  to  be  contained  in  A  II  three  times,  • 
and  in  A  C  five  times. 

arc  A  B 

llini  —  =  -• 

arc  A  C 

At  l:  on  A  B  and  A  C  draw  radii. 

These  radii  will  divide  Z  AOC  into  five  equal  parts,  of 

which  Z  A  0  B  will  contain  three,  §  180 

(jm  the  same  O,  or  equal  (D,  eqital  arcs  subtend  equal  A  at  the  centre). 

ZAOB_ 3 
'*'  Z  AOC       5* 

arc  AB 


Hut 


arc  A  C    '    5 

arc  A  B  __  Z  A  OB 

ztqAC  ~~  Z  AOC  ' 


Ax.  1. 


Q.  E.  D. 


92 


GEOMETRY. BOOK    II. 


Proposition  XIII.     Theorem. 

201.  In  the  same  circle,  or  in  equal  circles,  incom- 
mensurable arcs  have  the  same  ratio  as  the  angles  which 
they  subtend  at  the  centre. 

p,  P 


In  the  two  equal  ©  A  BP  and  A'B'P'  let  AB  and  A'  B 
be  two  incommensurable  arcs,  and  6T,  C  the  A  which 
they  subtend  at  the  centre. 

We  are  to  prove =  . 

1  arc  AB         AC 

Let  A  B  be  divided  into  any  number  of  equal  parts,  and 
let  one  of  these  parts  be  applied  to  A'  B'  as  often  as  it  will  be 
contained  in  A'li'. 

Since  AB  and  A' B'  are  incommensurable,  a  certain  num- 
ber  of  these  parts  will  extend   from   A'  to  some  point;  as  D, 
Laving  a  remainder  D B'  less  than  one  of  these  parte. 
Draw  C  I). 
Since  A  B  and  A'D  are  commensurable, 
arc  A'  h  _  Z  A'C'D 
arc  AS  ~~~  Z  ACB* 

{two  commensurable  ares  have  th<  to  as  the  A  which  tliey  subtend  at 

the  centre). 

Now  suppose  the  number  of  parts  into  which  A  B  is  divided 

to  be  continually  increased  ;  then  the  length  of  each  part  will 

become  leas  and  less,  and  the  point  I)  will  approach  nearer  and 

nearer  to  B',  that  is,  the  are  A'  J)  will  approach  the  arc  A1  W  as 

its  limit,  and  the  Z  A'C'D-the  Z  A'  <"  11'  as  itfl  limit. 


§200 


MEASUREMENT   OP   ANGLES.  93 

Then  the  limit  of  ?1*J>  will  be  arc  A' B  ,      ■ 
arc  A  B  arc  A  B 

and  the  limit  of  ^S~  will  be  Z  ^Lj?'. 
ZACB  ZACB 

Moreover,  the  corresponding  values  of  the  two  variables, 

namely, 

htcAB  ZACB 

are  equal,  however  near  these  variables  approach  their  limits. 

.'.  their  limits and are  equal       §  199 

arc  Z  AC  B  '  * 

Q.  E.  D. 


202.  Scholium.   An  a,  he  centre  is  said  to  be  meas- 

Tliis  expression  means  that  an  angle 

it  the  <  •  ntn   is  such  part  of  the  angular  magnitude  about  that 

point  (four  right  angles)  as  its  intercepted  arc  is  of  the  whole 

nicumferenoe. 

A  eiroiimferenoe  is  divided  into  3f>o  equal  arcs,  and  each 
died  a  degree,  denoted  by  the  symbol  (°). 

The  angle  at  the  centre  which  one  of  these  equal  arcs  sub- 
tanda  is  also  called  a  degree. 

A  quadrant   (one-fourth   a   circumference)  contains  there' 

90°  ;  and  a  right  angle,  subtended  by  a  quadrant,  con- 
ains  90°. 

Henoe  an  angle  of  30°  is  J  of  a  right  angle,  an  angle  of  45° 
is  J  of  a  right  angle,  an  angle  of  135°  is  $  of  a  right  angle. 

Thus  we  get  a  definite  idea  of  an  angle  if  we  know  the 
aumber  of  degrees  it  contains. 

A  degree  is  subdivided  into  sixty  equal  parts  called  min- 
ites,  denoted  by  the  symbol  ('). 

A  minute  is  subdivided  into  sixty  equal  parts  called  sec- 
axis,  denoted  by  the  symbol  ("). 


94 


GEOMETRY.  ■ 


•BOOK    II. 


Proposition  XIV.     Theorem. 

203.  An  inscribed  angle  is  measured  by  one-half  of  the 
arc  intercepted  between  its  sides. 

B  B  B 


Case  I. 

In  the  circle  PA  B  {Fig.  1),  let  the  centre  C  be  in  one 
of  the  sides  of  the  inscribed  angle  B. 

We  are  to  prove      Z  B  is  measured  by  \  arc  P  A. 

Draw  CA. 

CA  =  CB, 

(being  radii  of  the  same  O). 

.'.ZB  =  ZA,  §112 

(being  <>j>/><isi(r.  equal  sides). 

ZPCA=ZB  +  ZA.  §  105 

(the  exterior  A  of a  A  is  equal  to  the  sum  of  the  two  opposite  interior  A). 

Substitute  in  the  above  equality  Z  B  for  its  equal  Z  A. 

Then  we  have       ZPCA  =  2ZB. 

But  Z  PC  A  is  measured  by  A  P,  §  202 

(tlie  Z  at  the  centre  is  measured  by  the  intercepted  arc). 

.*.  2  Z  B  is  measured  by  A  P. 
.'.  Z  B  is  measured  by  J  A  P. 


MEASUKKMKN I    OF    ANCLES.  95 

Case  II. 

In    the    circle   BAM   (Fig:.    2),    let    the    centre    C    fall 
within  the  angle  E  B  A. 

We  are  to  prove      Z  E  B  A  is  measured  by  £  arc  E  A, 

Draw  the  diameter  BC  P. 

Z  PBA  is  measured  by  \  arc  PA,  (Case  I.) 

Z  P  B  E  is  measured  by  £  arc  P  E,  (Case  I.) 

.-.  ZPBA  +  ZPBEia  measured  by  £  (arc  PA  +  arc  P  E). 

.  .  Z  E  BA  is  measured  by  \  arc  E  A. 

Case  III. 

In  the  circle  B  FP  (Fig.  3),  let  the  centre  C  fall  with- 
out the  angle  A  B  /'. 

We  are  to  prove      Z  A  B  F  is  measured  by  £  arc  A  F. 
Draw  the  diameter  B  CP. 
Z  V  S  F  is  measured  by  J  arc  V  /'.  (Case  I.) 

Z  PBA  is  measured  by  £  arc  PA,  (Case  I.) 

.\Z  PBF-Z  PBA  is  measured  by  J  (arc  PF— arc  PA). 
.     A  I'>  F  v  h  2LTC  A  F. 

Q.  E.  O. 

204.  Corollary  1.     An  angle  inscribed  in  a  semicircle  is 

it  is  measured  by  one-half  a  semi-circumfer- 
hy  90°. 

205.  Con.  2.  An  angle  inscribed  in  a  segment  greater  than 
a  semicircle  is  an  acute  angle  ;  fof  it  is  measured  by  an  arc  less 
than  one  half  a  s.nii-circumference ;  i.  e.  by  an  arc  less  than  90°. 

206.  Cor.  3.  An  angle  inscribed  in  a  segment  less  than  a 
semicircle  is  an  obtuse  it  is  measured  by  an  arc  greater 
than  one-half  a  semi-circumference ;  i.  e.  by  an  arc  greater 
than  90°. 

207.  Cor.  4.    All  angles  inscribed  in  the  same  segment  are 
d,  for  tiny  are  measured  by  one-half  the  same  arc. 


96  GEOMETRY.  —  BOOK   II. 


Proposition  XV.     Theorem. 

208.  An  angle  formed  by  two  chords,  and  whose  vertex 
lies  between  the  centre  and  the  circumference,  is  measured  by 
one-half  the  intercepted  arc  pins  one-half  the  arc  intercepted 
by  its  sides  produced. 


Let  the  Z  AOC  be  formed  by  the  chords  A  B  and  CD. 
We  are  to  prove 

Z  A  0  C  is  measured  by  |  arc  AC  +  \  arc  B  D. 
Draw  A  D. 
ZCOA=ZD  +  ZAy  §105 

[the  exterior  Z.  of  a  A  is  equal  to  the  sum  of  tlie  two  opposite  interior  A  ). 

lint  Z  1)  is  measured  by  \  arc  A  C,  §  203 

(an  inscribed  Z  is  measured  by  \  the  intercepted  arc)  ; 

and  Z  A  is  measured  by  \  arc  B  Dy  §  203 

.'.  Z  C  0  A  is  measured  by  \  arc  A  C  +  J  arc  B  D. 

Q.  E.  D. 


Ex.  Show  that  the  least  chord  that  can  be  drawn  through 
a  ^iven  point  in  a  circle  is  perpendicular  to  the  diameter  drawn 
through  the  point. 


KEA8UBEM2NT    OP    ANGLES. 


97 


Proposition  XVI.     Theorem. 

209.  An   angle  formed  by  a  tangent   and  a  chord  is 
measured  by  one-half  the  intercepted  arc. 


Let  II  AM  be  the  angle  formed  by  the   tangent  OM 
and  chord  A  II. 

We  are  to  prove 

Z  II A  M  is  measured  by  \  arc  A  EH. 

Draw  the  dktaefei  A  OF. 

Z  FA  M  is  a  rt.  Z,  §  186 

'  a  tangent  at  tlve  point  of  contact  is  _L  to  it). 

A  F  A  Ifj  1). dng  a  rt.  Z,  Lb  I  by  \  the  semi-circura- 

Earen  ■»•  A  E  /•'. 

Z  FA  II  is  measured  by  £  arc  FH,  §  203 

(an  inscribed  Z  is  measured  by  i  the  intercepted  arc)  ; 

.-.  Z  FA  M  —  Z  FA  II  is  measured  by  \  (arc  A  EF—  arc  HF). 
.*.  Z  II A  M  is  measured  by  J  arc  A  EH. 

Q.  E.  O. 


98 


GEOMETRY.  —  BOOK   II. 


Proposition  XVII.     Theorem. 

210.  An  angle  formed  by  two  secants,  two  tangents,  or 
a  tangent  and  a  secant,  and  which  has  its  vertex  without  the 
circumference,  is  measured  by  one-half  the  concave  arc,  minus 
one-half  the  convex  arc. 
0 


M  D 

Fig.  1.  Fig.  2.  Fig.  3. 

Case  I. 
Let  the  angle  0  {Fig.  1)  be  formed  by  the  two  secants 
OA  and  OB. 

We  are  to  prove 

Z  0  is  measured  by  J  arc  A  B  —  £  arc  E  C. 
Draw  CB. 

ZACB  =  ZO  +  ZB,  §  105 

(the  exterior  /.of  a  A  is  equal  to  the  sum  of  the  two  opposite  interior  A  ). 

By  transposing, 

ZO  =  ZACB-ZB, 

But  Z  A  CB  is  measured  by  \  arc  A  B,  §  203 

(an  inscribed  Z  is  measured  by  {  the  intercepted  arc). 

ami  Z  B  is  measured  by  \  arc  C E,  §203 

. ' .  Z  0  is  measured  by  \  arc  A  B  —  \  arc  C  B. 


MEASUREMENT    OF   ANGLES.  99 

Case  II. 

Let  the  angle    0  (Fig.  2)   be  formed  by  the  two  tan- 
gents OA  and  OB. 

We  are  to  prove 

Z  0  is  measured  by  \  arc  A  MB  —  \  arc  A  SB. 
Draw  A  B. 

Z  ABC  =  Z  0  +  Z  OAB,  §  105 

{the  exterior  Zofa&is  equal  to  t/ie  sum  of  the  two  opposite  interior  A ). 

By  ti'ai; 

Z  0  =  ZABC-Z  OAB. 

I >ut       Z  ABC  is  measured  by  J  arc  A  M  ff,  §  209 

{an  /.formed  by  a  tangent  and  a  chord  is  measured  by  £  the  intercepted  arc), 

ami         Z  OA  B  is  measured  by  £  arc  A  SB.  §209 

.*.  Z  0  red  by  \  arc  A  M  B  —  h  arc  A  SB. 

Cask    III. 

Let  the  angle  0  {Fig.  3)    be   formed   by  the    tangent 
OB  and  the  secant  OA* 

We  are  to  prove 

Z  0  is  measured  by  \  arc  A  D  S  —  J  arc  C  E  S. 
Draw  CS. 

ZACS  =  Z  0  +  Z  CSO,  §  105 

(the  exterior  Zo/a  A  is  equal  to  the  sum  of  the  txoo  opposite  interior  A). 
By  transpn 

Z  0  =  Z  ACS-  Z  CSO. 

But         Z  ACS  is  measured  by  |  arc  A  D  S,  §  203 

{being  an  inscribed  /.). 

and         Z  CS 0  is  measured  by  £  arc  C  E S,  §  209 

{being  an  Z  formed  by  a  tangent  and  a  chord). 

.'.  Z  0  is  measured  by  \  arc  A  D  S  —  \  arc  C E S. 

Q.  E.  D. 


100 


GEOMETRY. BOOK  II. 


Supplementary  Propositions. 

Proposition  XVIII.     Theorem. 

211.    Two  parallel  lines    intercept    upon   the  circum- 
ference equal  arcs. 


Let  the  two  parallel  lines  CA  and  B  F  {Fig.  1),  inter- 
cept  the  arcs  C  B  and  A  F. 

We  are  to  prove      arc  C  B  =  arc  A  F. 

Draw  A  B. 

Z  A=Z  B,  §  68 

(being  alt. -int.  A  ). 

But  the  arc  CB  is  double  the  measure  of  Z  A. 
and  the  arc  A  F  is  double  the  measure  of  Z  B. 

.'.  arc  C  B  =  arc  A  F.  Ax.  G- 

Q.  E.  D. 

212.  Scholium.  Since  two  parallel  lines  intercept  on  tho 
circumference  equal  arcs,  the  two  parallel  tangents  M N  and 
OP  (Fig.  2)  divide  the  circumference  in  two  semi-circumferences 
AC  B  and  A  Q  By  and  the  line  A  B  joining  the  points  of  contact 
of  the  two  tangents  is  a  diameter  of  the  circle. 


SI rPPLKMKN TAIIV    PROPOSITIONS.  101 


Proposition  XIX.     Theorem. 

213.  If  the    su?n    of  two   arcs  be  less  than   a  cireum- 
:,<■<'  the  greater  arc  is  eubtended  by  He  greater  chord; 

a, id  CO%ver&  If,  the  greater  chord  .subtends  the  (/renter  arc, 

B 


p 

In   the  circle  A  CP  let   the  two  arcs  A  B  and  BC  to- 
gether   he  less   than   the   circumference,   and  let 
AB  be  the  greater. 
Wt  are  to  prove      chord  AB>  dwrd  B  C. 

Draw  A  C. 
In  the  A  ABC 

Z  C,  measured  by  £  the  greater  arc  A  B,         §  203 
is  greater  than  Z  A,  measured  by  \  the  less  are  B  C. 

.-.  tin-  Bide  .1  B  >  the  side  BC,  §  117 

,  a&the  greater  Z  has  the  greater  side  opposite  to  it). 

C(),  ,      U    the     chord    AB    be    greater    than    the 

chord  B  C. 

We  an  to  rrove      arc  A  B  >  arc  B  C. 

In  the  A  ABC,  ^  ^  „ 

ab>bc}  Hyp- 

r.ZOA,  §118 

(in  akthe  greater  Me  has  the  greater  Z  opposite  to  it). 

.-.  arc  A  A  double  the  measure  of  the  greater  Z  C,  is  greater 

than  the  ,i«-  BC,  fcuble  the  measure  of  the  lea:  Z  A.    ^  £  ^ 


102  GEOMETRY.  —  BOOK    II. 


Proposition  XX.     Theorem. 

21 1.  If  tM  nlrn'ofdwo  arcs  be  greater  than  a  circum- 
ference, the  greater  arc  is  subtended  by  the  less  chord ;  and, 
conversely,  the  less  chord  subtends  the  greater  arc. 

B 


E 

In  the  circle  BC  E  let  the  arcs  AECB  and  BA  EC 
together  be  greater  than  the  circumference,  and 
let  arc  AECB  he  greater  than  arc  B A  EC. 

If  V  are  to  prove      chord  A  B  <  chord  B  C. 

From  the  given  arcs  take  the  common  arc  A  E  C ; 

we  have  left  two  arcs,  C B  and  A  B,  less  than  a  circumference, 

of  which  0  B  is  the  greater. 

.'.  chord  C  B  >  chord  A  B,  $  213 

(when  the  sum  of  two  arct  it  less  than  a  circumference,  the  greater  arc  is 
sttbtcwt'tl  by  th<  greater  chord). 

.*.  the  chord  A  B,  which  subtends  the  greater  arc  AECB, 
is  less  than  the  chord  BC,  which  subtends  the  less  arc  BA  EC. 

Conversely  :  If  the  chord  A  B  be  less  than  chord  B  C. 

We  are  to  prove      arc  A  E  C  B  >  arc  B  A  E  C. 

Arc  A  B  4*  arc  A  E  C  B  =  the  circumference. 

Arc  BC  +  arc  B A  E C  =  the  circumference. 
.\  arc  AB+  arc  A  EC  B  =  arc  BC  +  arc  BAEC. 

Bat  arc  A  B  <  arc  B  C,  §  213 

(being  stibtended  by  the  less  chord). 

.'.arc  A  EC  B>  arc  BAEC. 

Q.  E.  D. 


CONSTRUCTION-.  103 


ON    CONSTRUCTInv 

Proposition  XXI.     Problem. 

2  15.    T<>  fiii (I  u  point  in  a  plane,  having  given  its  (lis- 
'ances  from  two  known  point* 

C 

• 


Let  A  and  B  be  the  two  known  points;  n  the  dis- 
tance of  the  required  point  from  A,  o  its  distance 
from  H. 

It  is  required  to  ft  'id  at  the  given  distances  from  A 

/:. 

From  A  as  a  centre,  with  a  radius  equal  to  ra,  describe  an  arc 

From  B  as  a  centre,  with  a  radius  equal  too,  describe  an  arc 

intersecting  the  fonnet  arc  at  C. 

C  is  the  required  point 

Q.  E.  F. 


216,  Corollary  1.   By  continuing  these  arcs,  another  point 

below  the  points  A  and  B  will  be  found,  which  will  fulfil  the 
conditions. 

2 1 7.  Cor.  2.   When  the  sum  of  the  given  distances  is  equal 
to  the  distance  between  the  two  given  points,  then  the  two  arcs 
described  will  be  tangent  to  each  other,  and  the  point  of  tan- 
will  be  the  point  required. 


104  GEOMETRY. BOOK    II. 

Let  the  distance  from  A  to  B  equal  n  +  o. 

From  A  as  a  centre,  with  a  \/ 

radius  equal  to  n,  describe  an  arc  ;  A'  c\  'B 

and  from  B  as  a  centre,  with  n 

a  radius  equal  to  o,   describe  an  n 

arc.  ~~ 

These  arcs  will  touch  each  ~ 

other  at  C,  and  will  not  intersect. 

.*.  C  is  the  only  point  which  can  be  found. 

218.  Scholium  1.  The  problem  is  impossible  when  the 
distance  between  the  two  known  points  is  greater  than  the  sum 
of  the  distances  of  the  required  point  from  the  two  given  points. 

Let  the  distance  from  A  to  B  be  greater  than  n  +  o. 

Then  from  A  as  a  centre, 
.with  a  radius   equal  to  n,  de-  ^'  '& 

scribe  an  arc; 


/  \ 

n 


and  ixom  .088  a  centre,  with  a 

radius  equal  to  o,  describe  an  arc. 
These  area  will  neither  touch 

.  0 

nor  intersect  each  other ; 

hence  they  can  have  no  point  in  common. 

219.  Scho.  2.  The  problem  is  impossible  when  the  distance 
between  the  two  given  points  is  less  than  the  difference  of  the 
distances  of  the  required  point  from  the  two  given  points. 

Let  the  distance  from  A  to  B  be  less  than  n  —  o. 

From  A  as  a  centre,  with  a  radius  ^^     "-^^x 

equal  to  n,  describe  a  circle  :  /  \ 

and  from  B  as  a  centre,   with  a      /  /  \    \ 

radius  equal  to  o,  describe  a  circle.  \  (     A  \    \ 

The  circle  described  from  B  as  a     \  \  /   / 

centre  will  fall  wholly  within  the  circle      \  \^  /    / 

described  from  A  as  a  centre;  o  \  / 

hence  they  can  have  no  poinl  in         n "^ ~~ 

common. 


CONSTRUCTIONS.  105 


Proposition 

XXII.     Problem. 

220. 

To  bisect 

a  given 

iht  line. 
C 

i\\ 

1 

B 

A 

1 

1 

.  1 . 

I 

Let  A  H  be  the  given  straight  line. 
I  to  bisect  the  line  A  B. 
From  A  and  B  as  centres,  with  equal  radii,  describ 

:     0   C 

Then  the  line  C  E  bisects  A  B. 
For,  (7 and  E,  being  two  points  at  equal  distances  from  the 
mitiefl  A  letennine  the  position  ofaJL  to  the  mid- 

dle point  otA  B.  {  <>0 

Q.  E.  F. 

PaOPOBITIOH   win.     Prohlbm. 
U  a  given  point   in  a  strain //f    /me,  to   erect   a 


!\ 


B 


*  11  0 

Let  0  be  the  given  point  in  the  straight  line  AB. 
It  is  required  to  erect  a  _L  to  the  line  A  B  at  the  point  0. 

Take  011=  OB. 
From  B  and   //  as  centres,  with  equal  radii,  describe  two 
arcs  intersecting  at  R. 

Then  the  line  joining  R  0  is  the  _L  required. 
For.  0  and  R  are  two  points  at  equal  distances  from  B  and  H,  and 
.*.  determine  the  position  of  a  J_  to  the  line  H B  at  its 
middle  point  0.  §  60 

Q.  E.  F- 


106 


GEOMETRY. BOOK  II. 


Proposition  XXIV.     Problem. 

222.  From  a  point  without  a  straight  line,  to  let  fall  a 
perpendicular  upon  that  line, 

C 


>i 


X 


II V 


'  K 


B 


Let  AB  be  a  given  straight  line,  and  C  a  given  point 
without  the  line. 

It  is  required  to  let  fall  a  A- to  the  line  A  B  from  the  point  C. 

From  C  as  a  centre,  with  a  radius  sufficiently  great, 

describe  an  arc  cutting  A  B  at  the  points  11  and  K. 

From  //  and  K  as  centres,  with  equal  radii, 

describe  two  arcs  intersecting  at  0. 

Draw  C  0, 

and  produce  it  to  meet  A  B  at  m. 

C  m  is  the  _L  required. 

For,  C  and  0,  being  two  points  at  equal  distances  from  // 
and  K,  determine  the  position  of  a  J_  to  the  line  UK  at  its 
middle  point.  §  60 

Q.  E.  F. 


CONSTRUCTIONS. 


107 


Proposition  XXV.     Problem. 

;.    T<>     instruct  an  arc  equal  to  a  given  arc  whose 
centre  is  a  given  point. 


C'< 


I    \ 


i    I 
B' 


Let  C  be  the  centre  of  the  given  arc  A  B. 

It  is  required  to  construct  an  arc  equal  to  arc  A  B. 

Draw  CB,  CA,  and  A  B. 

From  C  as  a  centre,  with  a  radius  equal  to  CB, 

describe  an  indefinite  arc  B'  F. 

From  B'  as  a  centre,  with  a  radius  equal  to  chord  A  B, 

describe  an  arc  intersecting  the  indefinite  arc  at  A'. 

Then  arc  A1  B'  =  arc  A  B. 

draw  chord  A'  B'. 


For, 


and 


The  (D  are  equal, 
(being  described  with  equal  radii), 

chord  A'  B'  =  chord  A  B ; 

.'.  arc  A'  B'  =  arc  A  B, 
(in  equal  (D  equal  chords  mbtend  equal  arcs). 


Cons. 
§  182 


Q.  E    F. 


108  GEOMETRY. BOOK   II. 


Proposition  XXYI.     Problem. 

224.  At  a  given  point  in  a  given  straight  line  to  con- 
struct an  angle  equal  to  a  given  angle. 

F 

\ 


B  B' 

Let  C  be  the  given  point  in  the  given  line  C  B1,  and 
C  the  given  angle. 

It  is  required  to  construct  an  Z  at  C  equal  to  the  Z  C. 

From  C  as  a  centre,  with  any  radius  as  C  B, 

describe  the  arc  A  B,  terminating  in  the  sides  of  the  Z. 

Draw  chord  A  B. 

From  C  as  a  centre,  with  a  radius  equal  to  C  B, 

describe  the  indefinite  arc  B'  F. 

From  B'  as  a  centre,  with  a  radius  equal  to  A  B, 

describe  an  arc  intersecting  the  indefinite  arc  at  A'. 

Draw  A'  C. 
Then  Z  <?'  =  Z  tf. 
For,  ]omA'B'. 

The  ©  to  which  belong  arcs  A  B  and  A1  B'  are  equal, 
(being  described  with  equal  radii). 

and  chord  A'  B'  =  chord  A  B ;  Cons. 

.-.  arc^l/i?/  =  arc^^,  §  182 

(in  equal  (D  equal  chords  subtend  equal  arcs). 

.-.ZC  —  ZC,  §180 

(in  equal  (D  equal  arcs  subtend  equal  A  at  tfw  centre). 

Q.  E.  F 


I' RUCTIONS. 


109 


Proposition  XXVII.     Problem. 
225.  To  bisect  a  given  arc. 


/X 


Let  AOB  be  the  given  arc. 

It  is  reqt<  <ect  tlu  arc  AOB. 

Draw  the  chord  A  B. 

From  A  ;in<l  B  as  centres,  with  equal  radii, 

describe  arcs  intersecting  at  E  and  (7. 

Draw  EC. 

E  C  bisects  the  arc  AOB. 

For,  K  and  (7,  being  two  points  at  equal  distances  from 
A  and  B,  determine  the  position  of  the  ±  erected  at  the  middle 
of  (herd  AB;  §  60 

and  a  _L  elected  at  the  middle  of  a  chord  passes  through 
the  centre  of  the  O,  and  bisects  the  arc  of  the  chord.  §  184 

Q.  E.  F. 


110 


GEOMETRY. BOOK   II. 


Proposition  XXYIII.     Problem. 
226.  To  bisect  a  given  angle. 


Let  A  E  B  be  the  given  angle. 

It  is  required  to  bisect  Z  A  E  B. 

From  E  as  a  centre,  with  any  radius,  as  E  A, 

describe  the  arc  A  0  B,  terminating  in  the  sides  of  the  Z. 

I  >r;iw  the  chord  A  B. 

From  A  and  B  as  centres,  with  equal  radii, 

describe  two  arcs  intersecting  at  C. 

Join  EC. 

E  C  bisects  the  Z  E. 

For,  E  and  C,  being  two  points  at  equal  distances  from  A  and 
7?,  determine  the  position  of  the  JL  erected  at  the  middle  of 
AB.  §  CO 

And  the  _L  erected  at  the  middle  of  a  chord  passes  through 
the  centre  of  the  O,  and  bisects  the  arc  of  the  chord.  §  184 

.*.  arc  A  0  z=  arc  0  B, 

.\Z  AEC  =  Z  BEC,  §  180 

(in  the  same  circle  equal  arcs  subtend  equal  A  at  the  centre). 

Q.  E.  F. 


CONSTRUCTIONS.  Ill 


Proposition  XXIX.     Problem. 

;.    Through  a  given  point  to  draw   a   straight   line 
parallel  to  a  given  straight  line. 

E _  // 


B 


Let  A  B  be  the  given  line,  and  II  the  given  point. 

It  is  required  to  draw  through  the  point  II  a  line  II  to  tJie 
X  B. 

Draw  II  A,  making  the  Z  II  A  B. 

At  the  point  H  construct  Z  A  II  B  =  Z  II A  B. 

Tl,  the  lm.   ////is  II  to  AB. 

Z  EH  A  -  Z  II A  B;  Cons. 

.-.///;  is  II  to  AB,  §69 

(when  two  straight  lines,  lying  in  the  same  plane,  are  cut  by  a  third  straight 
id.  A  be  equal,  the  lines  are  parallel). 

Q.  E.  F. 


Ex.   1.    Find  the  locus  of  the  centre  of  a  circumference  which 
-  through  two  given  pointa 

2.  Find  the  locus  of  the  centre  of  the  circumference  of  a 
given  radius,  tangent  externally  or  internally  to  a  given  cir- 
cumference. 

3.  A  straight  line  is  drawn  through  a  given  point  A,  inter- 
secting a  given  circumference  at  B  and  C.  Find  the  locus  of 
the  middle  point  P  of  the  intercepted  chord  BC. 


112  GEOMETRY. BOOK   II. 

^  Proposition  XXX.     Problem. 

228.  Two  angles  of  a  triangle  being  given  to  find  the 

third. 

R 

i 


E ^H 


Let  A  and  B  be  two  given  angles  of  a  triangle. 

It  is  required  to  find  the  third  A  of  the  A. 

Take  any  straight  line,  as  E  F,  and  at  any  point,  as  //, 

construct  Z  R II F  equal  to  A  J>, 

and  Z  SHE  equal  to  Z  A. 

Then  A  R II S  is  the  Z  required. 

For,  the  sum  of  the  three  A  at  a  A  «—  2  rt.  A,        §  98 

and  the  sum  of  the  three  A  about  the  point  77,  on  the  same 
side  of  FF=2  rt.  A.  §  34 

Two   A   of   the   A    being    equal    to    two   A   about    the 
point  //,  Cons. 

the  third  Z  of  the  A  must  be  equal  to  the  third  Z  about 
the  point  11. 


Q.  E.  F. 


CONSTRUCTIONS.  113 


Proposition  XXXI.     Problem. 

229.   Two  sides  and   the  included  angle  of  a  triangle 
being  given,  to  construct  the  triangle. 

M 

,.-Ac 

E 


Let  the  two   sides  of  the  triangle  be  E  and  F,  and 
the  included  angle  A. 

It  is  required  to  construct  a  A  having  two  sides  equal  to  E 
and  F  respectively \  and  their  included  Z  =  Z  A. 

Take  UK  equal  to  the  side  F. 

At  the  point  H  draw  tbe  Kne  Ht£% 

making  the  Z  K 11 M  —  Z  A. 

On  //  M  take  EC  equal  to  E. 

Draw  C  K. 

Then  A  C  H  K  is  the  A  required. 

Q.  E.  F. 


114  GEOMETRY. BOOK   II. 


Proposition  XXXII.     Problem. 

230.  A  side  and  two  adjacent  angles  of  a  triangle  being 
given,  to  construct  the  triangle. 

0 

/  \ 

E^- ^C 


Let  C E  be  the  given  side,  A  and  B  the  given  angles. 

It  is  required  to  construct  a  A  having  a  side  equal  to  C  E> 
and  two  A  adjacent  to  that  side  equal  to  A  A  and  B  respectively. 

At  point  C  construct  an  Z  equal  to  Z  A. 

At  point  E  construct  an  Z  equal  to  Z  B. 

Produce  the  sides  until  they  meet  at  0. 

Then  A  C  0  E  is  the  A  required. 

Q.  E.  F. 

231.  Scholium.  The  problem  is  impossible  when  the  two 
given  angles  are  together  equal  to,  or  greater  than,  two  right 
angles. 


CONSTRUCTIONS.  115 


Proposition  XXXIII.     Problem. 

232.   The  three  sides  of  a  triangle  being  given,  to  con- 
struct the  triangle. 

\C  m 


A^- 


B  o 

Let  the  three  sides  be  m,  u,  and  o. 

It  is  required  to  construct  a  A  having  three  sides  respectively, 
.  and  o. 

Draw  A  B  equal  to  n. 

From  A  as  a  centre,  with  a  radius  equal  to  o, 

describe  an  arc ; 

and  bom  B  as  a  centre,  with  B  radius  equal  to  m, 

describe  an  arc  intersecting  the  former  arc  at  C. 

Draw  CA  and  C  B. 

Then  A  C  A  B  is  the  A  required. 

Q.  E.  F. 

233.  Scholium.    The  problem  is  impossible  when  one  side 
nal  to  or  grtati  r  than  the  sum  of  the  other  two. 


116 


GEOMETRY. BOOK    II. 


Proposition  XXXIV.     Problem. 

234.   The  hypotenuse  and  one  side  of  a  right  triangle 
being  given,  to  construct  the  triangle. 


C 


Let  m  be  the  given  side,  and  o  the  hypotenuse. 

It  is  required  to  construct  a  rt.  A  having  tlie  hypotenuse 
equal  o  and  one  side  equal  m. 

Take  A  B  equal  to  m. 

At  A  erect  a  J_,  A  X. 

From  J^asa  centre,  with  a  radius  equal  to  o, 

describe  an  arc  cutting  A  X  at  C. 

Draw  C  B. 

Then  A  C  A  B  is  the  A  required. 

Q.  E.  F. 


CONSTRUCTIONS.  117 


Proposition  XXXV.     Problem. 

5.   The  base,  the  altitude,  and  an  angle  at  the  base, 
of  a  triangle  L>  n,  to  construct  the  triangle. 

:^ 


0 

Let  o  equal  the  base,  m  the  altitude,  and  C  the  angle 
at  the  base. 

It  is  required  tc  comitruct  a  A  having  the  base  equal  to  o, 
the  altitude  equal  to  m,  and  an  Z  at  the  base  equal  to  C. 

Take  A  B  equal  to  o. 

At  the  point  A,  draw  the  indefinite  line  A  R, 

making  the  Z  BAR  =  A  C. 

At  the  point  A,  erect  al^J  equal  to  m. 

From  X  draw  XS  II  to  A  B, 

and  meeting  the  line  A  R  at  S. 

Draw  SB. 

Then  A  A  SB  is  the  A  required. 

Q.  E.  F. 


118  GEOMETRY. BOOK    II. 


Proposition  XXXVI.     Problem. 

236.   Two  sides  of  a  triangle  and  the  angle  opposite  one 
of  them  being  given,  to  construct  the  triangle. 

Case  I. 


When  the  given  angle  is  acute,  and  the  side  opposite  to  it  is  less  tlmn 

the  other  given  side. 

D 

/ 

B/ 

Jh 

/  / 

\ 

/  / 
/  / 

\ 
\ 
\ 

/   / 

\ 

/    / 

\ 

\ 

%  /      / 

\ 

A  £-±i 

V--* 

C  \_        C         ^'  C" 

a/ 

a 

Let  c  be  the  longer  and  a  the  shorter  given  side,  and 
A  A   the  given  angle. 

It  is  required  to  construct  a  A  having  two  sides  equal  to  a 
and  c  respectively,  and  tlte  Z  opposite  a  equal  to  given  Z.  A. 
Construct  /.DAE  equal  to  the  given  Z.  A. 
On  AD  take  A  B  =  c. 
From  B  as  a  centre,  with  a  radius  equal  to  a, 
describe  an  arc  intersecting  the  side  A  E  at  C  and  C". 

Draw  B  C  and  B  C". 
Then  both  the  A  A  B  C  and  A  B  C"  fulfil  the  conditions, 
and  hence  we  have  two  constructions. 

When  the  given  side  a  is  exactly  equal  to  the  JL  B  C,  there 
will  be  but  one  construction,  namely,  the  right  triangle  ABC. 

"When  the  given  side  a  is  less  than  B  C,  the  arc  described 
from  B  will  not  intersect  A  E,  and  hence  the  problem  is  na- 
il >le. 


CONSTRUCTIONS. 


119 


Case  II. 

When  the  given  angle  is  acute,  right,  or  obtuse,  and  the  side  opposite 
to  it  is  greater  than  the  other  given  side, 
D 


A 


/ 


\     \/ 


fi 


Fi-    1. 


\     . 


^''C 


When  the  given  angle  is  obtuse. 
Construct  the  ADA  A'  (  Pig.  1)  equal  to  the  given  Z  S. 

Take  A  B  equal  to  a. 
From  J?  as  a  centre,  with  a  radius  equal  to  c, 
describe  an  arc  potting  EA  at  C,  and  EA  produced  at  C. 
Join  B  C  and  B  C. 

Then  the  A  A  B  C  is  the  A  required,  and  there  is  only  one 
construction ;  for  the  A  A  B  C  will  not  contain  the  given  Z  S. 

■  the  given  angle  is  acute,  as  angle  B  A  Cf. 
There  is  only  one  construction,  namely,  the  ABAC  (Fig.  1). 

Wkm  tJir  given  Z.  is  a  right  angle. 

There  are  two  constructions,  the  equal  A  B  A  C  and  B  AC 
(K$  2).  q.  E.  F. 

The  problem  is  impossible  when  the  given  angle  is  right  or 
obtaae,  if  the  given  side  opposite  the  angle  be  less  than  the 
other  given  side.  §  1 1 7 


1*20  GEOMETRY. BOOK   II. 


Proposition  XXXVII.     Problem. 

237.  Two  sides  and  an  included  angle  of  a  parallelo- 
gram being  given,  to  construct  the  parallelogram, 
R 


ftp- 

/! 

E 

/ 

/  ! 

/ 

/ 

/ 

/ 
/ 

/ 

1 

/ 

1 

/ 

1 

/ 

I 

— i 

A 

1 

in 

B 

1 

1 

0 

Let  m  and  o  be    the    two  sides,  and  C  the  included 
angle. 

It  is  required  to  construct  a  O  having  two  adjacent  ndei 

equal  to  m  and  o  respectively,  and  their  included  Z.  equal  to  Z  C. 

Draw  A  B  equal  to  o. 

From  A  draw  the  indefinite  line  AB, 

making  the  Z  A  equal  to  Z  C. 

On  AM  take  A  II  equal  to  m. 

From  H  as  a  centre,   with   a  radius  equal  to  o,  describe 
an  arc. 

From  B  as  a  centre,  with  a  radius  equal  to  m, 

describe  an  arc,  intersecting  the  former  arc  at  E. 
Draw  EH  and  E B. 
The  quadrilateral  A  B  E  H  is  the  CJ  required. 
For,  AB  =  HE,  Cons. 

A  11  =  BE,  Cons. 

.'.  the  figure  ABEH  is  a  O,  §  136 

(a  quadrilateral,  which  has  Us  opposite  sides  equal,  is  a  O ). 

Q.  E.  F. 


CONSTRUCTION.^.  121 


Proposition  XXXVIII.     Problkm. 

238.  To  describe  a  fence  through  three  points 

not  in  the  same  straight  line. 


/ 
/ 
/ 


\ 


A* 


0  \ 


^A 


<■'■ 


Let  the  three  points  be  At  B,  and  C. 
It  is  required  to  ference  through  tie  three 

po'tltt*     .1.      6 

iid  BC. 

Bisect  A  B  and  B  C. 

the  points  of  bisection,    E  and   /•',  erect  J§  intersect- 
a  0. 

m  0  as  a  nil  a  radius  equal  to  0  A,  describe  a 

circle. 

O  A  BO  is  the  O  required 

.  the  point  0,  being  in  the  _L  EO  erected  at  the  middle 
of  the  line  A  i  qua!  distances  from  A  and  B ; 

and  also,  being  in  the  _L  F  0  erected  at  the  middle  of  the 
line  C  B,  is  at  equal  dial  >m  B  and  C,  §  58 

{every  point  in  ///<•  _L  erected  at  the  middle  of  a  straight  line  is  at  equal 

'remities  of  that 

.'.  the  point  0  is  at  equal  distances  from  A,  B,  and  C, 
and  a  O  described  from  <>  as  a  centre,  with  a  radius  equal 
to  0  A,  will  [>ass  through  the  points  A,  B,  and  C. 

Q.  E.  F. 

.  SCHOLIUM,  The  same  construction  serves  to  describe 
a  circumference  which  shall  pass  through  the  three  vertices  of  a 
triangle,  that  is,  to  circumscribe  a  circle  about  a  given  triangle. 


122 


GEOMETKY. 


-BOOK    II. 


Proposition  XXXIX.     Problem. 

240.   Through  a  given  point  to  draw  a  tangent  to  a 
given  circle. 


Case   1.  —  When  tlie  given  point  is  on  the  circumference. 

Let  ABC  {Fig.  1)   be  a  given  circle,  and  C  the  given 
point  on  the  circumference. 

It  is  required  to  draw  a  tangent  to  the  circle  at  C. 

From  the  Centre  0,  draw  the  radius  OC. 
At  the  extremity  of  the  radius,  C,  draw  C  M  ±  to  0  C. 

Then  C  M  is  the  tangent  required,  §  186 

(a  straiglU  line  ±  to  a  its  extremity  is  tangent  to  tlic  O). 

Case  2.  —  Wlien  the  given  point  is  without  the  circumference. 

Let  ABC  (Fig.  2)   be    the  given  circle,  0  its  centre, 
£  the  given  point  without  the  circumference. 
It  is  required  i<>  draw  a  tangent  to  the  circle  ABC  from 

the  point  E. 

Join  0  E. 

On  0  E  as  a  diameter,  describe  a  circumference  intersecting 
the  given  circumference  at  the  points  M  and  11. 

Draw  0  M  and  0Ht  EM  and  EH. 

Now  /LOME  is  a  rt.  Z,  §  204 

{being  inscribed  in  a  semicircle). 

.'.  EM  is  _L  to  0  M  at  the  point  M \ 

/.EMU  tangent  to  the  O,  §  18G 

(a  straight  line  _L  to  a  radius  at  its  extremity  is  tangent  to  the  O). 
In  like  manner  we  may  prove  11  E  tangent  to  the  oiven  O. 

Q.  E.  F. 

241.  Corollary.  Two  tangents  drawn  from  the  same  point 

to  a  circle  arc  equal. 


CONSTRUCTIONS.  123 


Proposition     XL.     Problem. 
242.   To  inscribe  a  circle  in  a  given  triangle. 


\  J 

\ 

Let  ABC  be  the  given  triangle. 

-  ((O  m  the  A  A  B  C. 

iw  the  line  A  E,  bisecting  Z  A, 

and  draw  tin-  line  C  K,  bisecting  Z  C. 

w  /;//_!_  to  the  Mm  AC. 

From  A',  with  radios  AT/A  describe  the  O  K  M  If. 

the  O  K  II  M  La  the  O  required 

For,  «haw  A'A'i.  to^7i, 

and  A'  J/  _L  to  B  C. 

In  the  it.  A  A  A' A' and  i  // Af 

AE=A  A\  Iden. 

Z  BAK-Z  EAff,  Cons. 

.-.A  AKE  =  A  A  HE,  §110 

(7Vo  rf.  A  '  hypotenuse  and  an  acute  /.  of  the  one  be  equal 

respectively  to  the  hypotenuse  and  an  acute  Z.  of  the  other). 

.\  EK=.EII, 
(being  homologous  sides  of  equal  A). 

In  like  manner  it  may  be  shown  E M=  EH. 
.'.  EK,  EH,  and  E M are  all  equal. 
. ' .  a  O  described  from  E  as  a  centre,  with  a  radius  equal  to  E II, 

will  touch  the  sides  of  the  A  at  points  //,  K,  and  M,  and 
be  inscribed  in  the  A.  §  174 

Q.  E.  F. 


124 


GEOMETRY. BOOK  II. 


Proposition  XLI.     Problem. 

243.    Upon  a  given  straight  line,  to  describe  a  segment 
which  shall  contain  a  given  angle. 


H 


%. 


A^ 


\ 


AJt 


i 1 
/  i 
/  / 
/  / 


E 


7" 


Let  AB  be  the  given  line,  and  M  the  given  angle. 

It  is  required  to  describe  a  segment  upon  the  line  A  B,  which 
shall  contain  Z  M. 

At  the  point  B  construct  Z  A  BE  equal  to  Z  M. 
Bisect  the  line  A  B  by  the  1^^. 
From  the  point  B,  draw  B  0  _L  to  E  B. 
From  0,  the  point  of  intersection  of  F If  and  B  0,  as  a 
<(iitre,  with  a  radius  equal  to  OB,  describe  a  circumference. 

Now  the  point  0,  being  in  a  J_  erected  at  the  middle  of 
A  B,  is  at  equal  distances  from  A  and  B,  §  58 

(every  point  in  a  _L  erected  at  the  middle  of  a  straight  line  is  at  equal  dis- 
iancesfrom  the  extremities  of  that  line) ; 

.*.  the  circumference  will  pass  through  A. 
Now  B  E  is  J_  to  0  B,  Cons. 

.'.BE  is  tangent  to  the  O,  §  186 

(a  straight  line  ±  to  a  radius  at  its  extremity  is  tangent  to  the  O). 

.'.  Z  A  B E  is  measured  by  \  arc  A  B,  §  209 

tog  an  Z  formed  by  a  tangent  wnd  a  chord). 

Also  any  Z  inscribed  in  the  segment  A  II  li,  as  for  instance 
Z  A  A  /;,  Lb  measured  by  J  arc  AB,  §  203 

rlbi'l  Z). 


-  «.\^Ti:rcTiONS.  125 


.\Z  AKB  =  Z  ABE, 

(being  both  measured  by  £  the  same  arc)  ; 

.\Z  A  KB=*Z  M. 
gnient  J  //  B  La  the  segment  required. 


Q.  E.  F. 


Proposition   XLII.     Pbobudl 
244*    To  Jinil  the  ratio  of  two  commensurable  straight 

6   // 

A L-. LJ    B 


K 

CI , , r^D 

F 
Let  A  B  and  CD  be  two  straight  lines. 

It    M    r>  ,  /'    Of  A  II 

and  C  D,  so  as  to  express  tJu  ires. 

Apply  C '  D  to  A  B  as  many  times  as  possible. 

Suppose  twice  with  a  remainder  BB, 

Then  apply  KB  to  CD  as  many  times  as  possible, 

Suppose  three  times  with  a  remainder  F D. 
Th<u  apply  FD  t«>  BBbb  many  times  as  possihlc. 

Suppose  once  with  a  remainder  //  B, 
Then  apply  //  5  to  FD  as  many  times  as  possible. 

Suppose  once  with  a  remainder  AT  /A 

Then  apply  A'  />  to  //  B  is  many  times  as  possible. 

Suppose  A'/'  ined  jnsl  twice  in  HB. 

of  each  line,  referred  to  A"Z)  as  a  unit,  will 
than  be  as  follows  :  — 

//  B  =  '2  K  D  ; 

FD  =  HB+  K I)  •=  3A'Z>; 
BB  -  P2>  +  //A'  -  5A'j9; 
CD  =3EB+  FD  =  ISKD; 
AB  =2CD+  EB  =  il  KD. 

'  '  CD  =t    ISKD  ' 
...  the  ratio  of  —  --. 


126  geometry. book  ii. 

Exercises. 

1.  If  the  sides  of  a  pentagon,  no  two  sides  of  which  are 
parallel,  be  produced  till  they  meet ;  show  that  the  sum  of  all 
the.  angles  at  their  points  of  intersection  will  be  equal  to  two 
right  angles. 

2.  Show  that  two  chords  which  are  equally  distant  from  the 
centre  of  a  circle  are  equal  to  each  other ;  and  of  two  chords,  that 
which  is  nearer  the  centre  is  greater  than  the  one  more  remote. 

3.  If  through  the  vertices  of  an  isosceles  triangle  which  has 
each  of  tin*  angles  at  the  base  double  of  the  third  angle,  and  is 
in-ribed  in  a  circle,  straight  lines  he  drawn  touching  the  circle ; 
show  that  an  isosceles  triangle  will  be  formed  which  has  each 
of  the  angles  a*  one-third  of  the  angle  at  the  vertex. 

4.  A  I)  P>  is  a  semicircle  of  which  the  centre  is  C ;  and  A  EC 
is  another  Bemicircle  on  the  diameter  AC;  A  T  is  a  common 
tangent  to  the  two  semicircles  at  the  point  A.  Show  that  if 
from  any  point  /\  in  the  circumference  of  the  first,  a  straight 
line  FC  be  drawn  to  C,  the  part  F  K,  cut  off  by  the  second 
semicircle,  is  equal  to  the  perpendicular  F II  to  the  tangent  A  T. 

5.  Show  that  the  bisectors  of  the  angles  contained  by  the 
opposite  sides  (produced)  of  an  inscribed  quadrilateral  intersect 
at  right  angles. 

6.  If  a  triangle  A  BC  be  formed  by  the  intersection  of  three 
tangents  to  a  circumference  whose  centre  is  0,  two  of  which, 
A  M  and  AN,  are  fixed,  while  the  third,  BC,  touches  the  cir- 
cumference at  a  variable  point  P;  show  that  the  perimeter  of 
the  triangle  A  B  C  is  constant,  and  equal  to  AM  +  AN,  or 
'2  A  M.     Also  si iow  that  the  angle  B  0  C  is  constant. 

7.  A  B  is  any  chord  and  A  C  is  tangent  to  a  circle  at  A, 
C D  E  a  line  cutting  the  circumference  in  D  and  E and  parallel 
to  A  B ;  show  that  the  triangle  AC  J)  La  equiangular  to  the 
triangle  E  A  B. 


CONSTRUCTIONS.  127 


Constructions. 

I.  Draw  two  concentric  circles,  such  that  the  chords  of  the 
miter  circle  which  touch  the  inner  may  be  equal  to  the  diameter 
of  the  inner  circle. 

.  2.  Given  the  base  of  a  triangle,  the  vertical  angle,  and  the 
fcngth  of  the  line  drawn  from  the  vertex  to  the  middle  point 
of  the  base  :  construct  the  triangle. 

3.  Given  a  side  of  a  Mangle,  its  vertical  angle,  and  the  radius 
of  the  circumscribing  circle  j  construct  the  trial  -J  & 

4.  Given  the  base,  vertical  angle,  and  the  perpendicular  from 
the  extremity  p£  the  base  to  the  opposite  side  :  construct  the 
triangle.         1  ty  i 

5.  Describe  a  circle  cutting  the  aides  of  a  given  square,  so 
that  its  circumference  may  be  divided  at  the  points  of  inter- 
section into  eight  equal  arcs. 

le  of  60°,  one  of  30°,  one  of  120°,  one 
of  1")0°,  one  of  46p,  and  one  of  135°. 

7.  In  a  given  triangle  A  BC,  draw  Q  D  K  parallel  to  the  base 
BC  and  meeting  the  sides  of  the  triangle  at  J)  and  F,  so  that 
DEAv.xW  be  equal  to  DB  +  EC. 

8.  ( liven  two  perpendiculars,  A  B  and  CD,  intersecting  in  0, 
and  a  straight  line  intersecting  these  perpendiculars  in  i?and  F; 
to  001  one  of  whose  angles  shall  coincide  with 
one  of  the  right  anglee  at  0,  and  the  vertex  of  the  opposite  angle 
of  the  square  shall  lie  in  E  F.     (Two  solutions.) 

9.  In  a  given  rhombus  to  inscribe  a  square. 

10.  If  the  base  and  vertical  angle  of  a  triangle  be  given ; 
find  the  locus  of  the  vertex. 

II.  If  a  ladder,  whose  foot  rests  on  a  horizontal  plane  and 
top  against  a  vertical  wall,  slip  down ;  find  the  locus  of  its 
middle  point. 


BOOK   III. 

PROPORTIONAL  LINES  AND  SIMILAR  POLYGONS 


On  the  Theory  of  Proportion. 

24$.  Def.  The  Terms  of  a  ratio  are  the  quantities  com- 
pared. 

246.  I'ir.    The  Antecedent  of  a  ratio  is  its  first  term. 

247.  I  'ii.    The  Consequent  of  a  ratio  is  its  second  term. 

248.  Def.  A  Proportion  is  an  expression  of  equality  be- 
tween two  equal  ratios. 

A  proportion  may  be  expressed  in  any  one  of  the  following 
forms :  — 

1.  a  :  b  : :  e  i  d 

2.  a  :  b  *■  e  :  d 

3.  ?=c-. 
b       d 

Form  1  is  read,  a  is  to  b  as  c  is  to  d. 

Form  2  ia  read,  tin-  ratio  of  a  to  b  equals  the  ratio  of  c  to  d. 

Form  3  La  read,  a  divided  by  b  equals  <■  divided  by  cL 

Tin'  Terms  of  a  proportion  are  the  four  quantities  com- 
pare.]. 

The  first  and  third  terms  in  a  proportion  are  the  ante- 
cedents, the  second  and  fourth  terms  are  the  consequents. 

249.  The  Extremes  in  a  proportion  are  the  first  and  fourth 
terms. 

250.  The  Means  in  a  proportion  are  the  second  and 
terms. 


rHBOBT    OF   PROPORTION.  129 

251.  Dv,  In  the  proportion  a  :  b  :  :  c  :  d\  d  is  a  Fourth 
Pr<> i><> rti onal  to  a,  b,  and  & 

252.  Dip.  In  the  proportion  aibii'bic;  c  is  a  7%mi 
Proportional  to  a  and  6. 

25 3.  DBF.  In  the  proportion  a  :  b  :  :  b  :  c ;  b  is  a  Mean 
Proportional  between  a  and  c 

254.  Dkf.  Four  omantities  arc  Reciprocally  Proportional 
wh»n  tin-  lirst  is  to  the  second  as  the  reciprocal  o['  the  third  is  to 
tic  reciprocal  of  the  fourth. 

Thus  "  :  b  : :  —  : 

c     d 

11'  we  have  tw>  quant  od  />,  and  the  reciprocals  of 

and      :  these  ftrar  quantities  form  a  r>  <>?/>/■<>- 
a  b 

rii  proportion,  the  find   being  to  the  second  as  the  reciprocal  of 
the  second  ja  to  tie-  reciprocal  of  the  I 

1       I 

!>    :  :    -    :    -  . 
h      a 

255.  I  tar.    A  proporti  when  the 

i  ernes,  are  made  to  places. 

Tims  in  th«-  proportion 

a   :   b   :  :  <•   :   </. 
have  either 

a  :  :  '/.     or,     d  :  b  :  :  c  :  a. 

256.  Pi  i.  A  proportion  ifl  taken  hy  Inversion,  when  the 
ne-ans  and  extremes  are  made  to  exchange  places. 

Tims  in  the  proportion 

a  :  b  : :  c  :  d\ 

by  inversion  we  h 

b   :  a  :  :  d  :  c. 

257.  Dbp.    A  proportion  is  taken   by  Composition,  when 
urn  of  the  lirst  and  second  is  to  the  second  as  the  sum  of 


130  GEOMETRY. BOOK   III. 

the  third  and  fourth  is  to  the  fourth ;  or  when  the  sum  of  the 
tirst  and  second  is  to  the  first  as  the  sum  of  the  third  and  fourth 
is  to  the  third. 

Thus  if  a  :  b  :  :  c  :  d, 

we  have  by  composition, 

a  +  b   :  b   :  :  c  -r  d  :  d, 

or,  a  +  b  :  a  :  :  c  -f  d  :  c. 

258.  Def.  A  proportion  is  taken  by  Division,  when  the 
difference  of  the  first  and  second  is  to  the  second  as  the  dif- 
ference of  the  third  and  fourth  is  to  the  fourth ;  or  when  the 
difference  of  the  first  and  second  is  to  the  first  as  the  difference 
of  the  third  and  fourth  is  to  the  third. 

Thus  if  a  :  6  :  :  c  :  d, 

we  have  by  division 

a  —  b   :  b  :  :  c  —  did, 

or,  a  —  b  :  a   :  :  c  —  d  :  c. 

Proposition  I. 

259.    ///  every  proportion  ike  product  of  ike  extreme*  is 

equal  to  the  product  <>/'  the  means. 

Let  a  :  b  :  :  c  :  d. 

We  are  to  prove       a  d  ■■  b  & 


Now 

a 
b  = 

c 
d' 

whence, 

by  multiplying  by 

bd, 

ad  = 

■be. 

Q.  E.  D 


TIIKORY    OF    PROPORTION.  131 

In  the  treatment  ■  of  proportion,  it  is  assumed  that  fractions 
may  be  found  which  will  represent  the  ratios.  It  is  evident  that 
a  ratio  may  be  represented  by  a  fraction  when  the  two  quanti- 
ties compared  can  be  expressed  in  integers  in  terms  of  any 
common  unit.  Thus  the  ratio  of  a  line  2£  inches  long  to  a  line 
3£  inches  long  may  be  represented  by  the  fraction  Jg  when  both 
lines  an-  expressed  in  terms  of  a  unit  y1^  of  an  inch  long. 

But  it  often  happens  that  no  unit  exists  in  terms  of  which 
hotlt  th.'  quantities  can  be  expressed  in  integers.  In  such  cases, 
however,  it  is  possible  to  find  a  fraction  that  will  represent  the 
rutin  to  a  ■  d  degree  of  aeeura 

Thus,  if  a  an«l  B  denote  two  Lncommensarable  lines,  and  b  be 
divided  into  any  integral  number  (*)  of  equal  parts,  if  one  of 
these  parts  be  contained  in  a  more  than  m  tunes,  but  less  than 

//i-fl    times,    thru     "  >  ^L  but  < — — — ;    so    that   the    error 
b        n  it 

in    taking   either   of    these  values   for  -is  < -.     Since  n  can 

b  n 

be  increased  at  pleasure,  -  can  be  made  less  than  any  assigned 

value  whatever.    Propositions,  therefore,  that  are  true  for  —  and 

n 

m+ — ,  however  little  these  fractions  differ  from  each  other,  are 
n 

true  for  -  ;  and  _  may  be  taken  to  represent  the  value   of  -'. 
6  b 


Proposition  II. 

260.  A   mean  proportional   between    two   quantities   is 
equal  to  ike  square  root  of  their  product. 

In  the  proportion  a  :  b  :  :  b  :  c, 

b2  =  ac,  §259 

(the  product  of  the  extremes  is  equal  to  the  product  of  the  means). 

Whence,  extracting  the  square  root, 
b  =  ^  ac. 

Q.  E.  D. 


L32  GEOMETRY. BOOK    III. 

Proposition  III. 

261.  If  the  product  of  two  quantities  be  equal  to  the 
product  of  two  others,  either  two%/may  be  made  the  extremes 
of  a  proportion  in  which  the  other  two  are  made  the  means. 

Let  a  d  =  be. 

We  are  to  pt%ove      a  :  b  :  :  c  :  d. 

Divide  both  members  of  the  given  equation  by  b  d. 

Then  -  =  -, 

b        df 

or,  a  :  b  :  :  c  :  d. 

Q.  E.  D. 


Proposition  IV. 

12.    If  four  quantities  of  the  same  Hud  be  in  propor- 
h<>//,  they  will  be  in  proportion  by  alternation. 

Let  a  :  b  : :  e  :  d. 

We  are  to  prove      a  :  e  :  :  b  :  d . 

Now,  ^  =  £ 

h       ,1 

Multiply  each  member  of  the  eqiuition  by  -  . 

c 

Then  2  =  ±, 

c        d 

or,  a  :  c  :  :  b  :  d. 

Q.  E.  D. 


THEORY    OF    PROPORTION.  133 

Proposition  V. 

263.  If  four  quantities  be  in  proportion,  they  will  be  in 
proportion  by  inversion.      y* 

Let  a   :  b  :  :  c  :  d. 

We  are  to  prove     b  :  a   :  :  d  :  c. 

Now,  -  =  -  . 

h      (/ 

Divide  1  by  each  member  of  the  equation. 

Then  *  -  £ , 

a        c 


:  d  :  c. 


Proposition  VI. 


Q.  E.  D. 


quantities  be  in  proportion,  they  will  be  in 


proportion  by  composition. 


Let  a   : 

:  b  :  :  c  :  d 

a+8 

\b  :  :  c  +  d  : 

d. 

Now 

a 
b 

0 

Add  I  to  each  member  of  the  equation. 

Then 

a 

c 

that  is. 

a  +  b 

c+  rf 

or,  a  +  />   :  6   :  :  c  +  c/  :  d. 


q.  E.  d. 


134  GEOMETRY. BOOK   III. 

Proposition  VII. 

265.  If  four  quantities  be  in  proportion,  they  will  be  in 
proportion  by  division. 

Let  a  :  b  :  :  c  :  d. 
We  are  to  prove      a  —  b  :  b  :  :  c  —  d  :  d. 

Now  2«i. 

b        d 

Subtract  1  from  each  member  of  the  equation. 
Then 


that  18, 


a  c 

a  —  b        c  —  d 


b  d 

or,  a  —  b  :  b  :  :  c  —  d  :  d. 


Q.  E.  D. 


Proposition  VIII. 
266.    In  a  series  of  equal  ratios ,  the  sum,  of  the  ante- 
cedents  is  to  tie  sum  of  the  consequents  as  any  antecedent  is 
'<>  its  consequent. 

Let  a   :  b  =  c  :  d=e  :f  =  g  \  h. 
We  are  to  prove      a  +  c+e  +  gib  +  d-t-f+h:  :  a  :/>. 
Denote  each  ratio  by  r. 

a        c        e        g 
Then  r=b=d=f  =  h- 

Whence,     a  =  br,       c  =  drf       e=fr,       g  =  hr. 
Add  these  equations. 

Then        a  +  e  +  e  +  g  —  (6  +  d+/+  h)  r. 
Divide  by  (6  +  d  +  /  +  h). 

Then  a  +  c  +  e  +  y  a 

b+d+f+h      7        b 

or,     a  +  c  +  e  +  y  ;  b  -f  d  +  /  +  h  :  :  a  :  b. 

Q.  E.  D. 


THEORY    OP   PROPORTION. 


135 


:  c  : 

d> 

!  9 

K 

:  m 

n, 

fl   : 

:  cgm  :  dhn. 

9 

V 

k  _  M 
1        n  ' 

Proposition  IX. 

267.    The  products  of  the  corresponding  terms  of  two  or 

more  proportion*  are  in  proportion* 

Let  a  :  b 

e  :/ 
k  :  I 

We  are  to  prove      aek  :  bfl 

Now       I'r    / 

Whence  by  multiplied 

aek        e 

bfl         d  hn 

or,  ae('  :  6//  :  :  cam  :  <///  M. 

Q.  E.  D. 

1*1:0 POSITION    X. 

(8.   Like  powers,  or  like  roots,  of  the  terms  of  a  pro- 
port,  propori 

Let  a    :  b    :  :  t 

We  are  to  prove      an   :  bn   :  :  <n 

i_        1 

a»   :  A" 

Now 


I. 


:  d. 

1 


•  a        c 

b  =  d' 
By  raising  to  the  11th  power, 
cn 
6"  ~~  i? 
By  extracting  tin*  /'th  root, 


—  =  -    ;  or  a" 


1 
a* 


i 


tf\ 


rf* 


=  —  :  or,  an  :  b*  :  :cn 

1.  I 

60        eS 

Q.  E.  D. 

269.  Def.  Equimultiples  of  two  quantities  are  the  products 
obtained  by  multiplying  each  of  them  by  the  same  number. 
Thus  wirt  and  m  b  are  equimultiples  of  a  and  b. 


136  GEOMETRY.  —  BOOK    III. 


Proposition  XI. 

270.  Equimultiples  of  two  quantities  are  in  the  same 
ratio  as  the  quantities  themselves. 

Let  a  and  b  be  any  two  quantities. 
We  are  to  prove      ma  :  mb  :  :  a  :  b. 

xt  •  a        a 

Now  —  =  -  . 

b         b 

Multiply  both  terms  of  first  fraction  by  m. 

Then  n±a  =  1  , 

mb         b 

or,  ma  :  mb  :  :  a  :  b. 

Q.  E.  D. 

Proposition  XII. 

271.  //'  ttvo  quantities  be  increased  or  diminished  hi/ 
tike  parti  of  each,  the  result*  will  be  in  the  same  ratio  as  the 
quantities  tien$sehe$. 

Let  a  and  b  be  any  two  quantities. 

We  are  to  prove      a  ±  —  a  :  b  ±  -  b  :  :  a  :  b. 

9.  9 

In  the  proportion, 

ma  :  mb  :  :  a   :.b,    # 
substitute  for  m,  1  ±  -  . 


Then         (l  ±  t\  a  :  (\  ±  E\  b 

or  a  ±  P  a  :  b  ±  *-b  : 

7  7 


Q.  E.  D. 


272.  Def.    Euclid's  test  of  a  proportion  is  as  follows  :  — 
"The  first  of  four  magnitudes  is  said  to  have  the  same  ratio 
to  the  second  which  the,  third  lias  to  the  fourth,  when  any  equi- 
multiples whatsoever  of  the  first  and  third  being  taken,  and  any 
equimultiples  whatsoever  of  the  second  and  fourth  : 


THEORY    OF    PROPORTION.  137 

"  If  the  multiple  of  the  first  be  less  than  that  of  the  second, 
the  multiple  of  the  third  La  also  less  than  that  of  the  fourth;  or, 

"  If  fche  multiple  of  the  first  be  equal  to  that  of  the  second, 
Que  multiple  of  the  third  is  also  equal  to  that  of  the  fourth;  or, 

"If  the  multiple  of  the  fast  be  greater  than  that  of  the 
second,   the  multiple  of  the  third  is  also  greater  than  that  of 

the     t'uUltll." 


Proposition   XIII. 

;.   If  four  quantiUe$  be  proportional  according  to  the 
algebraical  definition,  iiejf  will  alio  be  proportional  according 

ti n  it ion. 

Let  a,  0,  r,  d   be  proportional  according  to  the  alge- 

a        c 
braical  definition  ;  that  is  7  =  -f  • 

h  d 

We  '  proportional  according  to  Euclid 's 

Multiply  each  member  [uality  by  —  . 


rra  ^  ma        mc 

nb  nd 

Now  from  the  nature  of  fractions, 

if  ma  be  less  than  »6,  m  c  will  also  be  less  than  nd ; 

it'  m  a  1"'  equal  to  n  bt  m  c  will  also  be  equal  to  nd ; 

if  m  a  be  greater  than  n  6,  m  c  will  also  be  greater  than  n  d. 

.'.  a,  6,  c,  (/  are    proportionals  according  to  Euclid's  def- 
inition. 

Q.  E.  D. 


138  geometry. book  iii. 

Exercises. 

x  1.    Show  that  the  straight  line  which  bisects  the  external 
vertical  angle  of  an  isosceles  triangle  is  parallel  to  the  base. 

2.  A  straight  line  is  drawn  terminated  by  two  parallel 
straight  lines;  through  its  middle  point  any  straight  line  is 
drawn  and  terminated  by  the  parallel  straight  lines.  Show  that 
the  second  straight  line  is  bisected  at  the  middle  point  of  the 
first 

3.  Show  that  the  angle  between  the  bisector  of  the  angle  A 
of  the  triangle  ABC  and  the  perpendicular  let  fall  from  A  on 
B  C  is  equal  to  one-half  the  difference  between  the  angles  B 
and  C. 

4.  In  any  right  triangle  show  that  the  straight  line  drawn 
from  the  vertex  of  the  right  angle  to  the  middle  of  the  hypote- 
nuse is  equal  to  one-half  the  hypotenuse. 

5.  Two  tangents  are  drawn  to  a  circle  at  opposite  extremities 
I   diameter,  and  cut  off  from  a  third  tangent  a  portion  A  B. 

If  0  be  the  centre  of  the  circle  show  that  A  CB  is  a  right  angle. 

6.  Show  that  the  sum  of  the  three  perpendiculars  from  any 
point  within  an  equilateral  triangle  to  the  sides  is  equal  to  the 
altitude  of  the  triangle. 

7.  Show  that  the  least  chord  which  can  be  drawn  through  a 
given  point  within  a  circle  is  perpendicular  to  the  diameter 
drawn  through  the  point. 

8.  Show  that  the  angle  contained  by  two  tangents  at  the 
extremities  of  a  chord  is  twice  the  angle  contained  by  the  chord 
and  the  diameter  drawn  from  either  extremity  of  the  chord. 

9.  If  a  circle  can  be  inscribed  in  a  quadrilateral  j  show  that 
the  sum  of  two  opposite  sides  of  the  quadrilateral  is  equal  to  the 
sum  of  the  other  two  sides. 

10.  If  the  sum  of  two  opposite  sides  of  a  quadrilateral  be 
equal  to  the  sum  of  the  other  two  sides;  show  that  a  circle 
can  be  inscribed  in  the  quadrilateral 


proportional  lines.  139 

On  Proportional  Lines. 

Proposition  I.     Theorem. 

274.  //'  a  series  of  parallels  intersecting  any  two 
si rn i <i hi  lines  intercept  equal  parts  on  one  of  these  lines, 
then  will  intercept  equal  parts  an  the  other  also. 

H  H' 


A  A' 

Let  the  series  of  parallels  A  A',  BB>,  CC,  DD\  EE', 
intercept  on  II' K'  equal  parts  A'  B'y  B'C,  CD1,  etc. 

We  are  to  pr< 

tic  If  K  equal  parts  A  B,  B  C,  CD,  etc. 

At  points  A  and  B  draw  A  m  and  Bn  II  to  W  K'. 

Am  =  A   B  .  §  135 

I  xmprehendrd  bet  wee  n  >>  re  equal). 

Bn  =  B'C',  §  135 

.*.  A  in  =  B  n. 
In  Hi.-  A  BA  m  and  0  Bn, 

ZA=ZB,  §  77 

{having  tf>-  iig  in  the  same  direction  from 

the  vertv 

Z  m  =  Z  n.  §  77 

and  A  m  =  B  //. 

.-.  ABAm  =  ACBn,  §  107 

[kavi  'mi  tirn  adj.  A  of  the  one  equal  respectively  to  a  side  and 

adj.  A  of  the  other). 

.\  AB  =  BC, 

[bt  mg  homologous  sides  of  equal  &). 
In  like  manner  we  may  prove  B C  =  CD,  etc. 


140 


GEOMETRY.  —  BOOK   III. 


Proposition  II.     Theorem. 

275.  If  a  line  be  drawn  through  two  sides  of  a  triangle 
parallel  to  the  third  side,  it  divides  those  sides  propor- 
tionally. 


.  1.  Fig.  2. 

In  the  triangle  A  B  C  let  E  F  be  drawn  parallel  to  B  C. 

We  are  to  prove        =     —  . 

*  AE        AF 

Case   I.  —  When  A  E  and  EB  {Fig.  1)  are  commensurable. 

Find  a  common  measure  of  A  E  and  BB,  namely  Bm. 

Suppose  Bm  to  be  contained  in  B  B  three  times, 

and  in  A  E  live  times. 

EB  =  3 

~AE        5* 


Then 


At  the  several  points  of  division  on  B  E  and  A  E  draw 
straight  lines  II  to  B  C. 

These  lines  will  divide  A  C  into  eight  equal  parts, 

of  which  FC  will  contain  three,  and  A  V  will  contain  five,    §274 

{if  parallels  intersecting  any  two  straight  lines  intercept  equal  parts  on  one 

of  these  I:  <  equal  parts  on  the  other  also). 

.    FC  _    3 
"  "  AF~~  5  ' 
EB       3 
AE 
.    EB  =  FC 
"  AE        AF 


But 


Ax.  1 


PROPORTIONAL    LINES.  141 


CASE.    II.  —  lllien  A  E  and  E  B  {Fig.  2)  are  incommensurable. 

Divide  A  E  into  any  number  of  equal  parts, 

and  apply  one  of  these  parts  to  E B  as  often  as  it  will  be 
contained  in  E  B. 

Since  A  E  and  E  B  are  incommensurable,  a  certain  number 
of  these  parts  will  extend  from  E  to  a  point  K,  leaving  a  re- 
mainder KB,  less  than  one  of  the  parts. 

Bmw  KH  II  to  BC. 

Since  A  E  and  E K  are  commensurable, 

:   „  =  -7—.  (Case  I.) 

Suppose  the  number  of  parte  into  which  A  E  is  divided  to 
be  continually  increased,  the  length  of  each  part  will  become  less 
and  lees,  and  the  point  K  will  approach  nearer  and  nearer  to  B. 

The  limit  of  E  K  will  be  E  Bs  and  the  limit  of  FH  will  be  FC 

.'.  the  limit  of     j—  will  be  , 

AE  AE 

and  the  limit  of  _ —  will  be  — — . 

J  /■  .  I  /■ 

Now  the  variables  —  and  — -  are  always  equal,  how- 
ever  near  they  approach  their  limits; 

.-.  their  limits  i-4  and  t-b  are  e(lual>  §  199 

Ah  A*  Q.  E.  D. 

BOLLABY.    One  side  of  a  triangle  is  to  either  part 
i -ut  off  1  hi  line  parallel  to  the  base,  as  the  other  side  is 

to  the  corresponding  part. 

Now  8  B  :  AE  ::  FC  :  A  F.  §275 

By  composition, 

A'  /;  +  A  /■    .  AE  :;  FC+  AF  :  AF,  §  263 

or,  AB  :  AE  ::  AC  :AF. 


142  GEOMETRY. BOOK   III. 


Proposition  III.     Theorem. 

277.  If  a  straight  line  divide  two  sides  of  a   triangle 
proportionally,  it  is  parallel  to  the  third  side, 

A 


In  the  triangle  A  BC  let  EF  be  drawn  so  that—  «._. 

AE     AF 

We  are  to  prove      E  F II  to  B  C. 

From  E  draw  E H  \\  to  BC. 

zi-jy  §276 

(one  side  of  a  A  wi  cut  off  by  a  line  II  to  the  base,  as  the  other 

side  is  to  the  corresponding  part). 

II -2$'         ' 

•    AC  -   A0  Ax  1 

.'.  AF  =  A  II. 
.'.  EF  and  JP  if  coincide, 

(their  extremities  being  the  same  points). 
But  E  II  is  II  to  B  C ;  Cons. 

.'.  EF,  which  coincides  with  EH,  is  II  to  BC. 

Q.  E.  D. 

278.  Def.   Similar  Polygons  are  polygons  which  havetheii 

homologous  angles  equal  and  their  homologous  sides  proportional. 
Homologous  points,   lines,  and  angles,  in  similar  polygons, 
are  points,  lines,  and  angles  similarly  situated. 


SIMILAR    POLYGONS.  143 


similar. 


On  Similar  Polygons. 
Proposition  IV.     Theorem. 
279.    Two  triangles  which  are  mutually  equiangular  are 


In  the  A  ABC  and  A' BO  let  A  A,  B,  C  be  equal  to 
A  A',  B'y  6"  respectively. 

IJ<  uretoprove      AB  :A'B'  =  AC  :  A'  C  =  BC  :  B'  C. 
Apply  the  A  A' B' C  to  the  A  ABC, 
so  that  Z  A'  shall  coincide  with  Z  A. 
Than  the  A  A'  B'C  will  take  the  position  of  A  A  E  H. 
Now  Z  A  E  II  |  ame  as  Z  £')  =  Z  B. 

i:  II  is  II  to  ^  (7,  §  69 

I  /ww  straight  lines,  ly  .  are  cw£  fa/  a  third  straight 

.  if  the  ext.  int.  A  be  equal  the  lines  are  parallel). 

AB  :  AE  -  AC  :  AH,  §276 

{one  side  of  a  A  is  to  off  by  a  line  II  to  the  base,  as  the  other 

side  is  to  the  corresponding  part). 

Suhetitute  for  A  K  and  A  H  their  equals  A'  B'  and  A'  C. 

Then  AB  :  A'B  =  AC  :A'C 

In  like  maimex  we  may  prove 

AB  :  A'B'  =  BC  :  B'C. 

.*.  the  two  A  are  similar.  §  278 

Q.  E.  D. 

280.  Cor.    1.    Two  triangles  are  similar  when  two  angles 
of  the  one  are  equal  respectively  to  two  angles  of  the  other. 

281.  Cor.  2.    Two  right  triangles  are  similar  when  an  acute 
of  the  one  is  equal  to  an  acute  angle  of  the  other. 


144  GEOMETRY. BOOK    ITT. 

Proposition  V.     Theorem. 

282.   Two  triangles  are  similar  when  their  homologous 
sides  are  proportional. 


In  the  triangles  A  BC  and  A'  B'  C  let 
A  ll       _A_C         M3 
ArB'  ~~  A'C1  ~"  B'C' 
We  are  to  prove 
A  A,  B,  and  C  equal  respectively  to  A  A',  B1,  and  C. 

Take  on  A  B,  A  K  equal  to  A'  B', 

and  on  AC,  All  equal  to  A'  C.     Draw  B  II. 

AB         AC  H 

A'B'  ~~  A'C'  m 

Substitute  in  tin's  equality,  for  A'  B'  and  A'  C  their  equals 
A  E  and  A  II. 

Then  ±1=    A(\ 

AE       AH 

.'.  Ellis  II  to  BC,  §  277 

Kif  „  Ifae  divide  two  sides  of  a  A  proportionally,  it  is  II  to  the  third  side). 

Now  in  the  A  ABC  and  A  E  If 

Z  ABC  =  Z  A  B  II.  §  70 

(being  cxt.  int.  angles). 

Z  AC B  =  Z  A  HE,  §  70 

Z  A=  Z  A.  Iden. 

.'.  A  A  B  0  and  A  Ell  axe  similar,  §  279 

(two  mutually  equiangular  &  are  similar). 

.    AB        A  S  ,  «,-„ 

(homologous  sides  of  similar  &  are  proportions  I). 


SIMILAR    POLYGONS. 


145 


But 


Since 


AB 
BC 
AE 
EH 


A'B1 
B'C* 
A'B' 
B'G' 


Hyp. 


Ax.  1 


Cons 


AE  =  A'B', 
EH  =  B'C. 

Now  in  the  A  A  EH  and  A'B'  C, 

EH  =  B'C,  AE  =  A'B'7  and  A  H  =  A'  C, 

.-.A  AEH  =  A  A'B'C,  §  108 

{having  three  sides  of  the  one  equal  respectively  to  three  sides  of  the  other). 

But  A  A  EH  is  similar  to  A  A  B  C. 

.'.  A  A'B' C  is  similar  to  A  ABC. 

Q.  E.  D. 

283.  Scholium.   The  primary  idea  of  similarity  is  likeness 
m  :  and  the  two  conditions  ueceesary  to  similarity  are  : 

I.  For  every  angle  in  one  of  tin    figures  there  must  bean 
equal  angle  in  the  other,  and 

II.  the  homologous  aides  must  be  in  proportion. 

In  the  case  of  triangle*  either  condition  involves  the  other, 
}>ut  in  the  case  of  S  it  does  not  follow  that  if  one 

condition  exist  the  pther  does  al 


Rt 


Thus  in  the  quadrilaterals  Q  and  Q',  the  homologous  sides 
are  proportional,  but  the  homologous  angles  are  not  equal  and 
the  figures  are  not  similar. 

In  the  quadrilaterals  R  and  B',  the  homologous  angles  are 
equal,  but  the  sides  are  not  proportional,  and  the  figures  are  not 
similar. 


146  GEOMETRY. BOOK   III. 

Proposition  VI.     Theorem. 

284.  Two  triangles  having  an  angle  of  the  one  equal  to 
an  angle  of  the  other,  and  the  including  sides  proportional, 
are  similar. 


A 


A' 


In  the  triangles  A  B  C  and  A'  B' C'  let 

£A=£  A',    and  —  =  i_£ . 
A'B'        A'C' 

We  are  to  prove      A  A  B  G  and  A'  B'  G'  similar. 

Apply  the  A  A'  B'  C  to  the  A  A  B  G  so  that  Z  A'  shaU 
coincide  with  Z.  A. 

Then  the  point  B'  will  fall  somewhere  upon  A  B,  as  at  E, 

the  point  C  will  fall  somewhere  upon  A  C,  as  at  H,  and 
B'C'uyonEH. 

xr  AB         AG  „ 

Now  = Hyp. 

A'B'       A'C  ™ 

Substitute  for  A'  B'  and  A'  C  fcheii  equals  A  E  and  A  H. 

Then  ±*_±*. 

AE       AH 

.'.the  line   Ell  divides  the  sides   AB  and  AC  propor- 
tionally ; 

.'.EHi*  II  to  BO,  §  277 

(if  a  line  divide  two  sides  of  a  A  proportionally,  it  is  II  to  the  third  side). 

.'.  the  A  A  2?Cand  A  EH  are  mutually  equiangular  and  similar. 

.'.A  A'  B'  G  is  similar  to  A  A  B  G. 

Q.  E.  D. 


SIMILAR   POLYGONS.  147 


Proposition  VII.     Theorem. 


285.    Two  triangles  which  have  their  sides  respectively 
parallel  are  similar. 


In  the  triangles  ABC  and  A1  B' C  let  AB,AC,    and 
BC   be  parallel  respectively   to  A' B',  A'C,   and 
B'  C. 
We  are  to  prove      A  ABC  and  A'  B'  C  similar. 

The  corresponding  A  are  either  equal,  §  77 

</?/•»>  A  whose  sides  arc  II,  in  the  same  direction,  or 

opposite  directions,  from  their  vertices  are  equal). 

or  supplements  of  each  other,  §  78 

iff  two  A  taw  tiro  sides  II  a  \  the  same  direction  from  their  vertices, 

frk  He  the  other  two  sides  are  II  and  lie  in  opposite  directions,  the  A  are 
fements  of  each  other). 

Hence  we  may  make  three  suppositions : 

1  >t.  A  +  A'  —  2  it  A,    B+B'  =  2Tt.A,     C+C'  =  2 rt.  A. 
2d  .1  =  1',  B+B'  =  2rt.A,     C  +  C  —  2 rt.  A. 

3d.  A=Af,  B  =  B'  .'.  C  =  C. 

Since  the  sum  of  the  A  of  the  two  A  cannot  exceed  four 
right  angles,  the  3d  supposition  only  is  admissible.  §  98 

.%  the  tvo  &  ABC  and  A'  B'  C  are  similar,         §  279 

(two  mutually  equiangular  &  are  similar). 

Q.  E.  D. 


148 


GEOMETRY. BOOK    III. 


Proposition  VIII.     Theorem. 

286.   Two  triangles  which  have  their  sides  respectively 
perpendicular  to  each  other  are  similar. 


In  the  triangles  E  F  D  and  B  A  C,  let  E F,  FD  and  ED, 
be  perpendicular  respectively  to  AC,  BC  and  AB. 

We  are  to  prove  A  E  F  D  and  B  AC  similar. 

Place  the  A  E FD  so  that  its  vertex  E  will  fall  on  A  />\ 
and  the  Bide  E  /•',  _L  to  A  C,  will  cut  A  C  at  /•". 

Draw  F'  1)<  II  to  F  1),  and  prolong  it  to  meet  BC  at  //. 
In  the  quadrilateral  B  ED  II,  A  E  and  //  are  rt.  A. 

.'.Z  B  +  Z  ED  II =2  rt.  A.  §158 

But  Z  E D>  F'  +  Z  ED  11=2  rt.  A.  §  34 

.\Z  ED'  F'  =  ZB.  Ax.  3. 

Now  ZC+  ZI/F'C=rt.Z,  §103 

(in  a  rt.  A  (lie  sum  of  the  two  acute  A  =  a  rt.  Z )  ; 

and              Z  K  F'  I)'  +  Z  II F'  C  =  rt.  Z.  Ax.  9. 

.'.ZEE'  D'  =  ZC.  Ax.  3. 

.'.  AE  F  D  and  BAC  are  similar.  §  280 

But              AEFD  is  similar  to  A  E F  D'.  §  279 

.*.  A  E FD  and  B  A  C  are  similar. 

Q.  E.  D. 

287.  Scholium.  When  two  triangles  have  their  sides  re- 
spectively parallel  or  perpendicular,  the  parallel  sides,  or  the 
perpendicular  sides,  are  homologous. 


si  MI  LAB   POLYGONS.  149 


Proposition  IX.     Theorem. 

288.  Lime*  drawn  through  the  vertex  of  a  triangle  divide 
proportionally  He  base  and  its  parallel. 


In  the  triangle  A  BO  let  II I  be  parallel  to  AC,  and 
let  BS  and  HT  be  lines  drawn  through  its  ver- 
tex to  the  base. 

We  <>ve 

As        ST        T_C 
WO  BB  OR       RL 

A  B  HO  and  BA  S  are  similar,  §  279 

(two  &  which  arc  mutually  equiangular  an-  similar). 

A  BO R  and  B ST  are  similar,  §  279 

A  B  I!  I  and  BTC  arc  similar,  §  279 

A  8  _  (SB\  _ll_(BT\=TC         §  278 
HO  ~"  \OB/  "  OR  "  \BRf        RL' 

(homologous  sides  of  similar  &  are  proportional). 


Ex.   Show  that,  if  three  or  more  non-parallel  straight  linos 
divide  two  parallels  proportionally,  they  pass  through  a  common 

point. 


150  GEOMETRY. BOOK   IH. 

Proposition  X.     Theorem. 

289.  If  in  a  right  triangle  a  perpendicular  be  drawn 
from  the  vertex  of  the  right  angle  to  the  hypotenuse : 

I.  It  divides  the  triangle  into  two  right  triangles  which 
are  similar  to  the  whole  triangle >  and  also  to  each  other. 

II.  The  perpendicular  is  a  mean  proportional  between 
the  segments  of  the  hypotenuse. 

III.  Each  side  of  the  right  triangle  is  a  mean  pro- 
portional between  the  hypotenuse  -and-  it*  adjacent  segment. 

\  IV.  The  squares  on  the  two  sides  of  the  right  triangle 
have  the  same  ratio  as  the  adjacent  segments  of  the  hypote- 
nuse. 

V.   The  square  on  the  hypotenuse  has  the  same  ratio  to 
the  square  on  fit  In  %  the  hypotenuse  has  to  the  segrrent 

adjacent  to  that  *'/<}<■. 
B 


In  the  right  triangle  A  BC,  let  B F  be  drawn  from  the 
vertex  of  the  right  angle  B,  perpendicular  to  the 
hypotenuse  A  C. 

I.    We  are  to  prove 

the  A  A  B  F,  A  B  C,  and  F  B  C  similar. 
In  the  rt.  A  BA  F  and  BA  C, 

the  acute  A  A  is  common. 

.*.  the  A  are  similar,  §  2-Sl 

{hro  rt.  &  are  similar  when  an  acute  Z  of  the  one  is  equal  to  an  acute  Z 
of  the  other). 

Inthert.  A  BCFa.n&BCA, 

the  acute  A  C  is  common. 

.'.  the  A  are  similar.  §  281 

Now  as  the  rt.  AABF  and  CBF  are  both  similar  to 
A  B  C,  by  reason  of  the  equality  of  their  z£, 

they  are  similar  to  each  other. 


SIMILAR   POLYGONS.  151 


II.  We  are  to  prove      AF  :  BF  :  :  BF  :  FC. 
In  the  similar  A  A  B  F  and  C  B  F, 

A  F,  the  shortest  side  of  the  one, 
B  F,  the  shortest  side  of  the  other, 
A'  /'.  the  medium  .side  of  the  one, 
F  C,  the  medium  side  of  the  other. 

III.  We  are  to  prove      AC  :  AB  ::  AB  :  A  F. 
In  tin-  similar  A  ABC  and  ji£/; 

^4  6',  the  longest  side  of  the  one, 

A  //,  the  longest  side  of  the  other, 

A  Bt  the  shortest  side  of  the  one, 

A  F,  the  shortest  side  of  the  other. 
Also  in  the  similar  A  A  BC  and  FB C, 

AC,  tin  longest  side  of  the  one, 
:   11  C,  the  longest  side  of  the  other, 
:  />'  (\  the  medium  aide  of  the  one, 
•.  FOt  the  medium  side  of  tin-  other. 

ra     m         4  AI?        AF 

IV.  \\  e  are  to  prove     ==  — — - . 

El?       FC 
In  the  proportion  A  C  :  A  ti  \  \  A  B  \  A  F, 

r&  =  A  C  X  A  I  \  §  259 

(the  }>i  he  extremes  is  tqiuil  to  the  product  of  the  means), 

and  in  tht'  proportion  AC  :  BC  : :  BO  :  FC, 

FT?=*ACX  FC.  §259 

Dividing  the  one  by  the  other, 

JTI?  _    ACX  AF 
2J7f  —  ACX  FC' 

Cancel  the  common  factor  A  C,  and  we  have 

r&  =  af 

BC*  "  ^C" 

V.  JJV  ,>>•£  fo  »?we     = 

J02        AF 
jTC2  =  ACXAC. 

IT?  =  iC'X^,  (Case  III.) 

Divide  one  equation  by  the  other ; 

then  ^?L  -  ACX  AC  -  — 

jf  "S1CX4^         ^7^'  Q.  E.  D. 


152  GEOMETRY. BOOK   III. 


Proposition  XI.     Theorem. 

290.   If  two  chords  intersect  each  other  in  a  circle,  their 
segments  are  reciprocally  proportional. 


Let    the    two    chords    AH   and  EF  intersect   at    the 
point  0. 

We  are  to  prove'    AO  :  EO  :  :  OF  :  OB. 

Draw  AF  an.  1  E  11. 

In  the  A  AO  Fund  FOB, 

Z  F=Z  /!,  §  203 

(each  being  measured  by  %  arc  A  E). 

Z  A  =  z  a;  §  203 

(each  being  measured  by  £  arc  F  B). 

.*.  the  A  are  similar.  ■     §  280 

&  are  similar  when  two  A  of  the  one  are  equal  to  two  A  of  tlvc  other). 

AVhence     A  0,  the  medium  side  <>f  the  one,  §  278 

:  EO,  the  medium  side  of  the  other, 
:  :  0  F,  the  shortest  side  of  the  one, 
:  OB,  the  shortest  side  of  the  other. 

Q.  E.  D. 


SIMILAR    POLYGONS.  153 


Proposition  XII.     Theorem. 

291,  If  from  a  point  without  a  circle  two  secants  be 
drawn,  the  whole  $ecanU  and  the  parts  without  the  circle 
are  reciprocally  proportional. 


Let  OB  and  OC  be  two  secants  drawn  from  point  0. 

<>/!    :    Or    :   :    <>  M    :    (t  II. 

Draw  EC  and  MB. 

In  the  A  o//r  :m,l  0JTB 

Z  0  is  common, 

Z  £  -  Z  C,  §  203 

(eacA  6y  ^  arc  H M). 

.'.  the  two  A  are  similar,  §  280 

are  /vy;/,//  Co  two  A  of  the  oilier). 

Whenoe      0  Ik  the  longest  side  of  the  one.  §  278 

OC,  the  longest  side  of  the  other, 
0  M.  the  shortest  side  of  the  one, 
0  //.  the  shortest  side  of  the  other. 

Q.  E.  D. 


154  GEOMETRY. BOOK    III. 


Proposition  XIII.     Theorem. 


292.  If  from  a  point  without  a  circle  a  secant  and  a 
tangent  he  drawn,  the  tangent  is  a  mean  proportional  between 
the  whole  secant  and  the  part  without  the  circle. 


0* 

Let  OB  be  a  tangent  and  0  C  a  secant  drawn   from 
the  point  0  to  the  circle  MBC. 

m  ore  to  prove      OC  :  0  B  :  :  0  B  :  0  M. 

Draw  5  if  and  B  C. 

In  the  A  0  2?  J/;. ml  OBC 

Z  0  is  common. 

Z  0  n  M  Lb  measured  by  \  arc  M  B,  §  209 

(being  an  £  formed  by  fl  tangent  and  a  chord). 

Z  C  is  measure.  1  by  |  arc  B  M,  §  203 

(be  in>j  <m  inscribed  Z. ). 

.\Z  0  BM  =  A  C. 

/.A  05(7  and  OBM  are  similar,  §  280 

{having  two  A  of  the  one  equal  to  two  A  of  the  other). 

Whence  OC,  the  longest  side  of  the  one,  §  278 

:  OB,  the  longest  ride  of  the  oilier, 

:  :  0  />,  the  shortest  side;  of  the  one, 

:  0  My  the  shortest  side  of  the  other. 

Q.  E.  D. 


SIMILAR    POLYGONS. 


155 


Proposition  XIV.     Theorem. 

).  If  two  polygons  be  composed  of  the  same  number 
of  triangle*  which  are  similar,  cork  to  each,  and  similarly 
placed,  then  the  polygon*  arc  simifor. 


In  the  two  polygons  ABODE  and  A'  B  C  J)' E\  let 
the  triangles  II  A  I'.  EEC,  and  C  E 1)  be  similar 
respectively  to  the  triangles  B'  A'  E\  B'  E' '  C\  and 
C  E  i>. 

Wk  on  to  jo-ove 
do  polygon  A  BC  If  E  rimilar  to  the  polygon  A'  VC'D'E*. 

Z  ,1  =  Z  A1,  §  278 

Z  ABE  =  Z  A'B'  E\  §278 

'    Z  EBC  =  Z  E'B'C,  §278 

Add  ill'-  last  two  equali 

Then  Z.  A  HE  A-  Z  E  BC =  Z  A'  B'  E'  -f  Z  E>  L' C ) 
or,  Z  A11C=Z  A'B'C. 

In  like  manner  we  may  prove  Z  BC D  =  Z  B'C  D\  etc. 
.*.  the  two  polygons  are  mutually  equiangular. 
N        A  E       A  B       (  EB\  _BC  _(EC\_CD  _  ED 
{ m  7J&  ~  IFF  "  \E'  B'l "~  B>  C      \E'  C')      C  D'     E'  &' 

milar  &  arc  proportional). 

.-.  the  homo]  lea  of  the  two  polygons  are  proportional. 

.*.  the  two  polygons  are  similar,  §  278 

{having  their  homologous  A  equal,  and  (heir  homologous  sides  proportional). 

Q.  E.  D. 


156 


GEOMETRY. BOOK    I  IT. 


Proposition   XV.     Theorem. 

:Ia\.  If  two  polygons  be  similar,  they  arc  composed  of 
Ike  same  nn  nber  of  triangles,  which  arc  similar  and  similarly 
placed. 


B  a  />" 

Let  the  polygons   A  BCDE  and  A  ll  C  D  A"  be  similar. 

m  tw<>  homologous  verl  E  and  Av, 

.haw  diagonal!  EBt  EC%  and  E'Bf,  P  <\ 

w,  m  i  .1  SB,  EBC,  EC  D 

similar  ///  to  A  .1   E  /;,  //  B  Q%   E  O  //. 

In  !l„.  .'.  .!  A'A'an.l  A1  IV  IV, 

ZJ-.  §  278 

(ft  ing  komolog  >  vr  pel  yens), 

a  E        a  B 


A'E'       A'B' 
(A.  ing  homologous  rides  of  simitar  polygons)* 

A  E  B  and  A'  E1  IV  are  Bimilar, 


5  278 


§  28 1 
tqiuU  i,»  an  /  of  the  oiher%  "and  the  including 
rides  proportional)* 

Z  ABC=Z  A'  IV  (\ 
\g  homologo  nilar  polygo 

A  A  BE  =  Z  A'  IV  E\ 
(being  homologous  A  of  similar 

,\  /  A  lie      /  A  It  E  =  Z  A'  IV  r  -  Z  A1  IV  IV. 
That  is  Z  EBC  —  Z  E'B'C. 


SIMILAR    POLTGOK8. 


157 


X(»\V 


also 


E  B         AB_ 

BB  ™  A'  l'>  ' 

l  o/  similar  &  ) ; 

BC   =   A  B 
~B'  ( 


A'B 

s  of  similar  polygons). 

.     E  B         BC^ 

i 


Ax.  1 

§284 


B'  C 

.'.  A  EBCvdA  E'B'&Bie  similar, 
(having  an  /L«jth>  on*  equal  to  an  /  "j  /!<>■  other,  and  tin   including  sides 

not). 

In  like  manner  w*  may  prove  A  E( '  D  -imilar  to  A  K'C D'. 


Q.  E.  D. 


\Y|.      Th» 


gon*  have  the 
ratio  as  any  two  horn 


i,  b  a 

Let  the  two  similar  polygons  be  A  BC  D  E  and  A' B'C  1)' E1 , 
and  let  J'  and  P  represent  their  perimeters. 

W<  on  to  pro*       /'-./'    ;  :  A  B  \  A'B'. 

A  B  i  A'B    :  :   nC  :  B'C  ::  CD  :  C D1  etc.     §  278 

qpns  ore  prvportionot). 

.-.  A  B  +  8(7,  etc.   :  A'B  +  BC,  etc  : :  j*  J  :  A'B,    §  266 
gruaJ  ra&M  //"  n»m  «////.  /nth,  sum  of  the 

rU). 


That  18 


/'   :   /"   :  :   J  B   :  A'B. 


Q.  E.  D. 


158  GEOMETRY. BOOK    III. 


Proposition  XVII.     Theorem. 

296.   The  homologous  altitude*  of  two  similar  triangles 
have  the  same  ratio  as  any  two  homologous  sides. 


In   the  two   similar  triangles   ABC   and  A'B'C,    let 
the  altitudes  be  BO  and  B'O'. 

w  BO  All 

We  on  t<>  I > rove       =  . 

n  o'      a1  h> 

In  the  it.  A  BOA  and  />'  OM', 

ZA=ZA'  §278 

urfogoua  A  of  the  similar  &  A  B  C  and  A'  Bf  C). 

.'.A  BOA  and  A  B'O'  A'  are  similar,  §  281 

Jim  ft,  A  having  an  acute  Z  <>/  the  one  equal  to  an  acute  Z.  of  tlie  other  are 

rimilar). 

.'.  their  homologous  sides  give  the  proportion 

BO         AB 

B'O'  ~  A'  ll1' 

Q.  E.  D 

207.  Cor.   1.    The  homologous  altitudes  of  similar  trian 
have  the  same  ratio  as  their  homologous  bfl 


SIMILAR    POLYGONS.  159 


In  the  similar  A  A  BO  and  A'  B1  C, 

AC         AB 

A'C        A' B'' 

§278 

{the  homologous                imHar  &  are  proportional). 

And  in  tin'  similar  A  BO  A  and  B'  0'  A', 

BO         AB 
B'O'  ~    A'B1' 

§290 

.BO         AC 

&  0'  "  A1  C  ' 

Ax.  1 

;.  •_'.   Tin-  homologous  altitudes  <>i'  similar  triangles 

have  th.  io  as  their  perimeti 

Denote  the  perimeter  of  the  firs!   by  /\  and  thai  of  the 
second  bj  /' . 

Then  ''  §  295 

P      A'  /;  s 

(///'   jwrimctcrs  of  two  similar  polygons  have  the  same  ratio  as  any  two 
homologous  sides), 

Hut  —  =   *JL  j  §296 

Ax.  1 


Ex.  1.  It  any  two  straight  lines  be  cut  by  parallel  lines, 
Bhow  that  the  corresponding  segments  are  proportional. 

l\  [f  the  f  m  sides  of  any  quadrilateral  be  bisected,  show  that 
the  lines  joining  the  points  of  bisection  will  form  a  parallelo- 
gram. 

3.  Two  circles  intersect;  the  line  AH  KB  joining  their 
centres  A,  B,  meets  them  in  //,  K.  On  A  B  is  described  an 
equilateral  triangle  ABCy  whose  sides  BC>  A  C,  intersect  the 
circles  in  /•',  /'.  FE  produced  meets  BA  produced  in  P.  Show 
that  •  1 1  P  K  so  is  C  F  to  CE,  and  so  also  is  PH  to  PB. 


BO 

AB 

B  o 

A'B1' 

BO 

P 

wo 

=  /"' 

160  GEOMETRY. BOOK    III. 


Proposition  XVIII.     Theorem. 

299.  In  any  triangle  the  product  of  two  sides  is  equal 
to  He  product  of  the  segments  of  the  third  side  formed  by  the 
bisector  of  the  opposite  angle  together  with  the  square  of  the 
bisector. 


\  7 

\        / 
i: 

Let  Z  B  A  0  of  the  A  ABC  be  bisected  by  the  straight 
line  A  7A 

Wi  ore  to  prove      B  A  X  A  C  =  B 1)  X  D  C  +  A  1)\ 
Describe  the  O  A  B  C  about  the  A  A  B  C ; 
produce  A  1)  to  meet  the  circumference  in  E,  and  draw  EC 
Then  in  the  AABD  and  A  EC, 

Z  BA  D  —  Z  CAE,  Hyp. 

Z  B  =  Z  E,  §  203 

(ea< ,  flu  arc  AC). 

..AABD  and  A  EC  are  Bimilar,  $  280 

'.nrr  similar  when  two  A  of  the  our  are  equal  respectively  to  two  A 

Whence        B  A.  the  longest  ride  of  the  one, 
E  A,  the  longest  Bide  of  the  other, 
A  /k  the  shortest  side  of  the  one, 
A  ( ',  tli<'  shortest  side  of  the  other ; 

BA        A  D  s  0*fl 

or,  §  2/8 

AM         ilr  * 

(//"//"  i  <>f  si ui ilu r  k\  are  proportional). 

.'.BAX  AC  =  EAX  A  J). 
But  AM  X  Ah      {ED  +  .4  Z>)  J  0, 

.'.  7ivi  X  A  C  -  AT7>  X  yl  />  +  i(  J)\ 

But  E  DX  AD  =  BDX  h(\  §290 

(/Ar  segments  of  two  chords  in  a  G  ujft&ft  intersect  each  other  an 
reciprocally  proportional). 

Substitute  in  the  above  equality  />  />  X  DC  For  A'  />  X  A  D, 


then  BA  X  AC  =  BDX  DC  +  A  h\ 


Q.  E.  D. 


SIMILAR     POLYGONS. 


1G1 


Proposition  XIX.     Theorem. 

300.  In  any  triangle  tie  product  of  two  sides  is  equal  to 
the  product  of  the  diameter  of  ike  circumscribed  circle  by  the 
perpendicular  let  Jail  upon  tie  third  tide  from  the  vertex  of 

pposite  angle. 

- —    ~~^.<4 


/ 

Let   ABC    be  a  triangle,   and  AD    the  perpendicular 
from  A  to  EC. 

IVsrril.i'  tin-  ci iv u inference  A  BO  about  the  A  A  BC. 

uneter  A  A\  and  draw  S 

II     tm  t  .prove      BA  X  A  C  =  KA  X  A  D. 

In  the  &ABI>  and  A  EC 

Z  B  D  A  isa  it.  Z, 

Z  /  a  rt  Z, 

£  BDA  =Z  EC  A. 

Z  B  =  Z  A\ 

|  </w;  «rc  A  C). 

.*.  A  .!  /;  h  and  .1  #(7  aw  similar, 

(/„-,,,/..  nute  Z  of  tlie  other  are 

AM.  the  longest  side  of  the  one, 
AM,  the  Longest  side  of  the  other, 
A  h,  the  shortest  side  of  the  one, 
.1  ( '.  tin'  shortest  side  of  the  other; 

BA    =  AD 
£~A~  AC' 

\  BA  X  AC  =  EAX  AD. 


Cons. 
§204 

§203 
§281 


Win 


or, 


§278 


Q.  E.  D. 


162 


GEOMETRY. BOOK  III. 


Proposition  XX.     Theorem. 

301.  The  product  of  the  two  diagonals  of  a  quadrilateral 
inscribed  in  a  circle  is  equal  to  the  sum  of  the  products  of  its 
opposite  sides. 

C 


Let  ABC  D  be  any  quadrilateral  inscribed  in  a  circle, 
AC  and  B  1)  its  diagonals. 

Wt  are  to  pmr,       B  I)  X  A  C  =  A  B  X  C  f)  +  A  D  X  B  C. 

n  tmct  Z  A  BE  =  Z  DBC, 

and  add  to  each  Z  E  B  D. 

Then  iii  the  A  A  /if)  ;nu\  BOB, 

Z  ABD  =  Z  CBBt  Ax.  2 

and  Z  BDA=  Z  BC  E,  §203 

]>  i»  inn  measurnf  by  |  the  arc  A  B). 

.'.AABD  and  B  C  E,  are  similar,  §  280 

(two  &  are  similar  when  two  A  of  the  one  are  equal  respectively  to  two  A 
of  the  other). 

Whence       A  />,  the  medium  aide  of  the  one, 
C  E,  the  medium  side  of  the  other, 
B  D,  the  longest  side  of  the  one, 
B  C,  the  longest  aide  of  the  other, 


SIMILAR    POLYGONS. 


1G3 


or, 


AD  _    BD 
~CE  "  TU' 
(the  homologous  iidet  of  similar  &  arc  proportional). 


§  278 


.-.  BD  X  C  E  =  AD  X  BC. 
Again,  in  the  A  A  B  E  and  B  C  1>, 

Z.  ABE  =  Z  DBC, 
and  ZBAE  =  ZBDC, 

B  C). 


Cons. 
§203 

§  280 


§  278 


.*.  A  A  B  E  and  BC D  are  similar, 
</•  wkt  ii  two  A  of  the  one  are  equal  respectively  to  two  A 
;<  other). 

Whence       A  Ii  the  l  le  <•{'  the  one, 

/;  Ik  the  Longest  Bide  of  the  ot] 
A  /•;.  the  shorted  ride  of  the  one, 
0  l>.  ii;.  •  ride  i  f  the  other. 

A  li       A  /■: 
/;/>  =  CD> 

(//,,  &  are  proportional). 

.-.  /;  //X  AE=ABX  CD. 
Bui  />'/>  XCJS=  ADX  BC. 

Adding  these  two  equalities, 

BD  (A  E  +  f  E)  =  i  5  X  CD+  A  DX  BC, 
or  BDXAC  =  ABXCD  +  ADXBC. 

Q.  E.  D. 


Kx.    If  two  circles  are  tangent  internally,  show  that  chords 
0f  tl  :.  drawn  from  the  point  of  tangency,  are  divided 

proportionally  by  the  circumference  of  the  less. 


164  GEOMETRY. BOOK    III. 


On  Constructions. 
Proposition  XXI.     Problem. 
302.   To  divide  a  given  straight  line  into  equal  parts. 

A^r, , 7 -.B 


^0 


Let  A  11  be  the  given  straight  line. 

It  is  required  to  dimdt  A  B  into  equal  parts. 

From  A  draw  the  indefinite  line  A  0. 

'lake  any  convenient   length,  ami  apply  it  to  A  0  as  many 
times  as  the  line  A  11  is  t<>  be  « 1  i \  i*  1*  *  1  into  i>;irls. 

in  tlic  Lasi  point  thus  Pound  on  A  0,  as  C,  draw  OB, 

Through  the  several  points  of  division  on  A  0  draw  lines 
II  to  OB. 

These  lines  divide  A  B  into  equal  parts,  §  274 

(if  a  scries  of\\s  intersecting  any  two  straight  lines,   intercept  equal  parts 
on  one  of  these  lines,  they  intercept  equal  parts  on  tlie  other  also). 

Q.  E.  F. 


Ex.   To  draw  a  common  tangent  to  two  given  circles. 

I.  When  the  common  tangent  is  exterior, 

II.  When  the  common  tangent  is  interior. 


CONSTRUCTIONS.  165 


Proposition  XXII.     Problem. 

803,    To  tit  rule  a  given   -straight  line  into  part*  pro- 
portional  In  any  number  of  given  lines. 

\  \  I!  K B 


\ 


Let  A  /  and  o  be  given  straight  lines. 

If  U  required  to  '•  B  into  parti  proportional  to  the 

I ':  iw  the  indefinite  line  A  X. 
On  A  X  lake  AC^m, 

and  /;/'  =  o. 

*  /•'  /;.     Prom  E  and  C  draw  E  K  and  C II  II  to  FB. 
AT  and  ffare  the  division  points  required 

F,,  (±X)=±Z=  **=**,  §275 

\AJ:f        AC        GE        EFJ 

[a  li  through  two  rides  of  a  A  II  fo  thr  third  ride  divides  those 

,-f  i,  >,)<(! 1 if). 

.-.  A  If  :  UK  :  K  n  ::  AC  :  CE  :  E F. 
Substitute  >/>,  n.  and  o  for  their  equals  ^4  (7,  C  E,  and  7?./^ 
Then  J  //  :  HK  \  KB  ::  m  :  n  :  a 

Q.  E.  F 


166  GEOMETRY.  BOOK    III. 


Proposition  XXIII.     Problem. 

30i.    To   find   a  fourth    proportional   to    three    given 

straight  lines. 

B                           F                    m 
A^~ .  


5  >-   R 


Let  the  three  given  lines  be  m,  n,  and  o. 

It  m  /  to  find  a  fourth  proportional  to  m,  ft,  and  o. 

Take  A  B  equal  to  n. 

Draw  the  indefinite  line  A  R,  making  any  convenient  Z. 
with   A  R 

On  A  B  take  A  C  =  m,  and  C  S  =  o. 

Draw  C  B. 

Prom  8  draw  S  F  II  to  (7.5,  to  meet  A  B  produced  at  F. 

B F  is  the  fourth  proportional  required 

For,  AC  :  AB  :  :  CS  :  BF,  §  275 

(a  ton  '//■»"/■,/  through  two  ridu  of  a  A  II  to  the  third  side  divides  those  aides 
jrroportioiially). 

Substitute  to,  >/,  and  o  for  their  equals  i  (7,  i  B9  and  C  & 
Then  m  :  n  ::  o  :  BF. 

Q.  E.   F. 


CONSTRUCTIONS.  167 


Proposition  XXIV.     Problem. 
305.   To  find  a  third  proportional  to  two  given  straight 
A 


lines 


a 

B 

bL 

— \c 

' 

\ 
\ 

A 

c 

/ 

\ 
\ 
\ 

/ 

/ 

>*' -  - 

\ 
\ 

\ 
\ 

/? 

■ 

Let  A  B  and  AC  be  the  two  given  straight  lines. 

It  is  required  to  find  a  third  proportional  to  A  B  and  A  C. 

Place  A  B  and  A  C  so  as  to  contain  any  convenient  Z. 

Produce  A  B  to  A  making  BD^  AC. 

Join  BC. 

Through  J)  draw  J)  K  II  to  BC  to  meet  A  C  produced  at  E. 

C  E  is  a  Quid  proportional  to  A  B  and  A  C.        §  251 

£!-^,  §275 

(d  line  drawn  through  two  sides  of  a  A  II  to  Hie  third  side  divides  those  sides 
proportionally). 

Substitute,  in  the  above  equality,  A  C  for  its  equal  B  D ; 

Then  AB  =  AC 

AC        CE 

or,  A  B  :  A  C  :  :  A  C  :  CE. 

Q.  E.  F. 


1G8  GEOMETRY. BOOK   III. 

Proposition  XXV.     Problem. 
306.   To  find  a  mean  proportional  between    two  given 


lines. 


n 

7?\ 


'•  b ::b E 

Let  the  two  given  lines  be  m  and  n. 
It  if  required  to  find  a  mean  proportional  between  m  and  n. 
On  the  straight  line  A  B 

take  AC  —  m,  and  ('  B  =  n. 

On  A  B  as  a  diameter  describe  a  semi-circumference. 

At  C  erect  the  _L  0  II. 

C  II  is  a  mean  proportional  between  m  and  » 

Draw  //A' and  II  A. 

The  Z  ,1  //  /;  is  a  rt.  Z,  §  204 

(beiny  inscribed  in  a  flemia're/e), 

and  HClB  a  J.  l«'t  i'all  from  the  vertex  of  a  it.  Z  t<>  the 
hypotenuse. 

.-.AC  :  Off  ::  OH  :  C/i,  §289 

(2fa  ±  let  fall  from  0  •'//><  /•/.  Z  /"  //"   hypotenuse  is  d  mean  pro- 

portional between  tin-  segments  of  the  hypotenuse). 

Substitute  for  A  0  and  0  B  their  equals  m  and  ??.. 

Tla-n  m   :   (77/  :  :  C  If  :  ??.  Q   E   F 

307.  Corollary.  If  from  <t  /><>int  in  the  circumference  a 
perpendicular  be  drawn  to  dt<j  diameter,  <n<<l  chords  from  the  point 
to  the  extremities  of  the  diameter,  the  perpendicular  is  <>  nxan  pro- 
portional between  tin-  segments  of the  diameter,  and  each  chord  is  << 

*  proportional  between  Us  >i<ij<tm,t  segment  and  tl<<  diameter* 


CONSTRUCTIONS.  169 


Proposition  XXVI.     Problem. 

308.    To   divide  one  side  of  a  triangle  into  two  parts 
proportional  to  the  other  two  sidee. 


B JF 

Let  A  BC  be  the  triangle. 

It  u  BC  into  two  such  parti  that 

trig  shall  tqwal  the  ratio  of  the  other  two 
AC  and  A  B. 

dnce  0  A  to  /•',  making  .1  /  ■■  A  B. 
Draw  F II. 
A  draw  A  B     to  FB. 
the  division  point  required. 

For  *LA  =  rA.  $  275 

(a  lint  "V  '  h  two  sides  of  a  AW  to  the  0  ividea  those  sides 

Substitute  fox  A  F  its  equal  A  B. 

CA        C  E 


Thru 


A  B        E  B 

Q.  E.  P- 


300.  Cobollart.    The  line  A  F  bi sects  the  angle  CA  B. 
Foi  ZF=ZABF.  §112 

inal  sides). 

ZF=ZCAE,  §70 

.-lilt.    A  ). 

Z  ABF=  Z  BAE,  §68 

///  alt. -int.  A  ). 

.-.  Z  CA  F  =  Z  BAE.  Ax.  1 

310.    Dbf.    A  Btraight  line  is  said  to  be  divided  in  extreme 
and  mean  ratio,  when  the  whole  line  is  to  the  greater  segment 
segment  is  to  the  less. 


170 


GEOMETRY. BOOK    III. 


Proposition  XXVII.     Problem. 
311.   To  divide  a  given  line  in  extreme  and  mean  ratio. 


/ 
I 

E 


\D 


V 


A- 


/ 


u         s 

Let  A  B  be  the  given  line. 

If  £i  required  to  dividx  A  l>  in  extreme  and  mean  ratio. 

At  /;  «•!«•<•[  a  _L  11  ( ',  equal  to  one-half  of  A  B. 

From  0  aa  a  centre,  with  a  radius  equal  to  CBf  describe  a  O. 

Since  A  B  ia  .1.  to  the  radius  CB  at  its  extremity,  it  la 
tangent  to  the  circle. 

Through  C  draw  A  D,  meeting  the  circumference  in  E  and  D. 

On  J  B  take  d  //  =  d  A'. 
//  ia  tin- division  point  of  A  B  required. 


For 


A  1)    :    A  B    :  :    A  B    :    A  £ 


§  292 

row  a  pafttf  iriijnnit  the  circumfi  >  iU  and  a  tangent  be  drawn, 

the  tangent  is  a  mean  proportional  "between  tic  whole  Meant  and  the  part 
without  the  circumference). 


Then     A  D  —  A3  :  AB  :  :  A  8  -  A  E  :  d  E. 


§  268 


CONSTRUCTIONS.  171 


Since  AB='2CB,  Cons. 

and  ED  =  2CB, 

{the  diameter  of  a  Q  being  tu-ice  tlie  radiiis), 

AB  =  ED.  Ax.  1 

■\  A  I)  -  A  B  =  A  1)  -  ED  =  A  E. 
But  A  E  =  A  II,  Cons. 

.-.  .1  I)-  A  11  =  A  II.  Ax.  1 

A  II-  A  E  =  A  H-  A  11  =  II B. 
Substitute  these  equivalent  in  the  last  proportion. 

Then  A  II    :    A  B   :  :   II  B   i    A  II. 

Whence,  by  inversion,  A  B  \  A  II  \  ■  A  II  \  II B.     §  2G3 

.' .  A  B  is  divided  at  //  in  extreme  and  mean  ratio. 

Q.  E.  F. 

IiKmakk.  A  B  is  said  to  be  divided  at  //,  internally,  in 
extreme  and  mean  ratio,  [f  BA  be  produced  to  //',  making 
J  //  equal  to  AD,  A  />'  is  said  to  be  divided  at  II',  externally, 
in  extreme  and  mean  ratio. 

A  B    :    ■!  //     :   :    .!  //     :    //    //. 

When  a  line  is  divided   internally  and  externally  in  the 
is  said  to  be  divided  hgrmonioaUy. 

ThvmJB    '     f       f £  is   divided  harmoni- 


cally at  r;md  /A  if  (7  J  :CB::DA:DB;  that  is,  if  the  ratio 
of  the  distances  of  0  from  -1  and  />  is  equal  to  the  ratio  of  the 
distances  pf  I)  &om  .1  and  B. 

This  proportion  taken  by  alternation  gives: 

AC  :A  D  ::B  C  :  B  D  ;  that  is,  C  Z>  is  divided  harmoni- 
cally  at  the  points  B  and  it.  The  four  points  A,  B,  C,  D,  are 
called  hat  **U  ;  and  the  two  pairs  A,  B,  and  C,  D,  are 

called  cot[  '■■>?.<. 


Ex.  1.  To  divide  a  given  line  harmonically  in  a  given  ratio. 
2.  To  find  the  locus  of  all  the  points  whose  distances  from 
iveii  points  are  in  a  given  ratio. 


172 


GEOMETRY. 


■BOOK    III. 


Proposition  XXVIII.     Problem. 

812.    Upon  a  given  line  homologous  to  a  given  side  of  a 

give  a  polygon,  to  construct  a  polygon  similar   to   the  given 
polygon. 


B  C 

Let  A'  E'  be  the  given  line,  homologous  to  A  E  of  the 
given  polygon  ABC  D  E. 

It  is  required  to  On  A'  E'  a 'polygon  similar  to  the 

polygon. 

From  A  draw  the  diagonals  A' A' and  EC. 

in  E>  -haw  /•/  /; ,  making  Z  A'  P  11'  =  Z.I  A  II. 

Also  from  A!  draw  A1  A",  making  Z  W  A1  V  =  Z  BAE, 

and  meeting  E1  B'  at  B'. 

The  two  A  A  B  A  and  i'  &  E>  are  similar,  §  280 

'.  ore  similar  if  they  have  two  A  of  the  one  equal  respectively  to  two  A 
of  the  oth 

Also  tr-  .in  Av  draw  Av  C",  making  Z  A'  JP  C  =  A  BEG. 

From  A"  draw  ff  <?,  making  Z  A7  AT"  -  Z  A  AC, 

and  meeting  A'  6"  at  C 

Then  the  fcwo  A  A  Btf  and  A'  A'  C  are  similar,      $  280 
(te>  A  arc  similar  if '///. <//  tow  feao  d  o/£fo  ojm  0?uaJ  respectively  to  two  A 
of  the  other). 

In  like  mannei  conatruct  A  A'  c  /)'  similar  to  A  ECD. 

Then  the  two  polygons  are  similar,  §  293 

(top  polygons  composed  ofthesanu   number  of  &  similar  to  each  other 
similarly  placed,  are  similar). 

.'.  A' B'  C  I)'  E'  La  the  required  poly 

Q.  E.  F. 


i:\ercises.  173 


Exercises. 

1.  ABC  is  a  triangle  inscribed  in  a  circle,  and  B  D  is  drawn 
to  meet  the  tangent  to  the  circle  at  A  in  D,  at  an  angle  ABB 
equal  to  the  angle  ABC;  show  that  A  C  is  a  fourth  propor- 
tional  t<>  the  lines  A'  />,  A  D,  A  B. 

2.  Show  tliat  cither  of  the  sides  of  an  isosceles  triangle  is  a 
mean  proportional  between  the  base  and  the  half  of  the  segment 
of  the  base,  produced  if  necessary,  which  is  cut  off  by  a  straight 
line  drawn  from  the  vertex  at  right  angles  to  the  equal  side. 

3.  A  B  ifl  the  diameter  of  ;i  circle,  J)  any  point  in  the  circum- 
ttoe,  and  G  the  middle  point  of  the  arc  AD.     If  A  C,  A  D, 

BC  he  joined  and  .1  D  cut  BC  in  E,  show  that  the  circle  cir- 
cumscribed about  tie*  triangk  .1  EB  will  touch  AC  and  its 
diameter  will  he  a  third  proportional  to  BC  and  A  II 

4.  From  the.  ob1  le  of  a  triangle  draw  a  line  to  the 
base,  which  shall  be  a  mean  proportional  between  the  segments 
into  which  it  divid 

Find  the  poinl    in  the  oduced  of  a  right  triangle, 

fn»m  which  the  line*  drawn  to  the  angle  opposite  to  the  base 
shall  have  the  same  ratio  to  the  base  produced  which  the  per- 
icular  has  t.>  the  base  itself. 
G.  Aline  touching  two  (in  les  cuts  another  line  joining  their 
centres;  Bhow  that  the  segments  of  the  latter  will  be  to  each 
other  as  the  diameters  of  the  circles. 

7.  1  the  locus -of  the  middle  points  of  all  the  chords 
of  a  circle  which  pass  through  a  fixed  point.  : 

8.  0  ifl  point  from  which  any  straight  line  is  drawn 
meeting  a  lix  lit  line  at  P;  in  OP  a  point  Q  is  taken 
such  that  0  Q  is  to  0  P  in  a  fixed  ratio.  Determine  the  locus 
of  Q. 

9.  0  is  a  fixed  point  from  which  any  straight  line  is  drawn 
meeting  the  circumference  of  a  fixed  circle  at  P  ;  in  0  P  a  point 

taken  such  that  0  Q  is  to  OP  in  a  fixed  ratio.     Determine 

the  1 


BOOK  IV. 


COMPARISON    AND    MEASUREMENT    OF    THE    SUR- 
FACES  OF  POLYGONS. 

Profosition  I.     Theorem. 

313.   Tico  rectangles  having  equal  altitudes  are  to  each 
other  as  their  bases. 


D 


D 


B 


Let   the  two  rectangles  be  A  G  and  A  F,  having  the 
the  same  altitude  A  D. 


Wk  "re  to  prove 


rect.  A  G 
rect.  A  F 


AB 
AF' 


Case   I.  —  When  A  B  and  A  E  are  commensurable. 
Find  a  common  divisor  of  the  bases  A  B  and  A  F,  as  A  0. 
Suppose  A  0  to  be  contained  in  A  B  seven  times  and  in 


A  F  four  times. 

Then 


A  B       7 
A  F  ~  4  ' 

At  the  several  points  of  division  on  A  B  and  A  E  erect  _k . 
The  rect.  A  C  will  be  divided  into  seven  rectangles, 
and  rect.  A  F  will  be  divided  into  four  rectangles. 

These  rectangles  are  all  equal,  for  they  may  be  applied  to 
each  other  and  will  coincide  throughout. 


But 


rect  A  G 

7 

rect  A  F 

4 

AB 

AF  ~ 

7 
4' 

rect  A  G 
rect  A  F 

AB 
AF 

OOMPABISON  AVI)  MEASUREMENT  OF  POLYGONS.    175 


II.  —  //  'hen  A  B  and  A  E  are  incommensurable. 


D 


D 


H 


B 


K 


E 


Divide  A  B  into  any  number  of  equal  parts,  and  apply  one 
of  these  parts  to  .1  B  as  often  as  it  will  be  contained  in  A  E. 

Since  A  B and  A  B  are  incommensurable,  a  certain  number 
of  these  parts  will  extend  bom  .1  to  a  point  K,  leaving  a  re- 
mainder K  B  less  than  one  of  these  parts. 

Draw /T// II  to  7: '/. 

sine.'  .1  /;  and  .1  A' aiv  commensurable, 
not  AH       AK 


Case  1 


rect.  A  C 

the  nomber  of  parts  into  which  d  B  is  divided  to 
be  continually  increased,  the  Length  of  each  part  will  become  less 
and  lees,  and  the  point  K  will  approach  nearer  and  nearer  to  E. 

The   limit    of  A  K  will  be  d  /:,',  and  the  limit  of  rect.  All 
will  be  rect  .1  /•'. 

.-.the  limit  of  —  will  be  ^L?, 
AJi  AB' 

1  ,,     ,       .     f  rect.  d  //     .n   .     rect.  ^4  /" 
and  the  hunt  ol    r-r=   will   be 


rect  .1  ( ' 


rect.  A  C 


tables    -     -  and   — —      are    always    equal 

AB  rect  AG  J      H 


howei  'hey  approach  their  limits; 

rect.  A  F 


.*.  their  limits  are  equal,  namely, 


rect.  A  C 


AE 
~ABJ 


§199 


Q.  E.  D. 


31  d.  Corollary.  Two  rectangles  having  equal  bases  are 
to  each  other  as  their  altitudes.  By  considering  the  bases  of 
these  two  rectangles  A  D  and  A  D,  the  altitudes  will  be  A  B  and 
A  E.  But  we  have  just  shown  that  these  two  rectangles  are  to 
each  other  as  A  B  is  to  A  E.  Hence  two  rectangles,  with  the 
B  base,  or  equal  bases,  are  to  each  other  as  their  altitudes. 


176 


GEOMETRY.  —  BOOK  IV. 


Another  Demonstration. 
Let  A  C  and  A'  C  be  two  rectangles  of  equal  altitudes. 
PC  O  P> 


F        E        D        A  A'  D>  E'  F'  G> 

rect.  AC         AD 


We  are  to  prove 


rect.  A'  C        A'  D' 


Let  b  and  &',  S  and  S'  stand  for  the  bases  and  areas  of  these 
rectangles  respectively. 

Prolong  A  D  and  A'  //. 

Take  A  D,  D  E,  E F  .   .  .  .   m  in  number  and  all  equal, 

and  A'  D',  D'  E',  E'  F',  F  G'  .  .  .  .  //  in  number  and  all  equal. 

Complete  the  rectangles  a-  in  the  li«;ure. 

Then  base  A  F  =  m  b, 

and  base  A' G'  =  nb' ; 

rect.  AP  =  mS, 
and  rect  A'  P  =  n  S'. 

Now  we  can  prove  by  superposition,  that  if  A  F  be  >  A'  G', 
rect.  A  P  will  be  >  rect.  A'  P' ;  and  if  equal,  equal ;  and  if  less, 
less. 

That  is,  if  m b  be  >  nb1,  mS  is  >  n S1 ;  and  if  equal, 
equal  j  and  if  less,  less. 


Hence, 


6   :  V  :  :  £  :  S',         Euclid's  Prf.,  §  272 

Q.  E.  D. 


I  oMPAItlSON    AND    MEASUREMENT    OF    POLYGONS.        177 


iiv,;.»vs  177 


UmVERSI 


Proposition  II.     Theorei 


£&mvM 


3 1 5.    Ztao  rectangles  are  to  each  other  as  the  products  of 
Heir  bate*  by  their  altitude*. 


R> 


a' 


J 


b  V  b 

Let  R  and  R'  be  two  rectangles,  having  for  their  bases 
•  b  and  £',  and  for  their  altitudes  a  and  a', 
#         a  X  b 

R'  " 


We  are  to  prove 


and 


§314 


§313 


</'  X  // 

Construd  the  rectangle  Si  with  its  base  the  same  as  that 
i.l  its  altitude  the  same  as  that  of/?7. 

Then 

tangles  having  the  same  base  are  to  each  other  as  their  altitudes) ; 

£       ft 

IV  "  V ' 
{rectangb  dude  are  to  each  other  as  their  bases). 

By  multiplying  these  two  equalities  together 

R  aX  b 

Tl>  ™  d'Xfr*' 

Q.  E.  D. 

316.  Def.   The  Area  of  a  surface  is  the  ratio  of  that  surface 
to  another  surface  assumed  as  the  unit  of  measure. 

317.  Def.    The  Unit  of  measure  (except  the  acre)  is  a  square 
a  side  of  which  is  some  linear  unit ;  as  a  square  inch,  etc. 

318.  Def.    Equivalent  figures  are  figures  which  have  equal 
areas, 

E&HM.   In    comparing   the   areas   of  equivalent   figures   the 
symbol  ( =  )  is  to  be  read  "equal  in  area." 


178 


GEOMETRY. BOOK   IV. 


Proposition  III.     Theorem. 

319.  The  area  of  a  rectangle  is  equal  to  the  product 
of  its  base  and  altitude. 


U 


b  1 

Let  R  be  the  rectangle,  b  the  base,  and  a  the  alti- 
tude ;  and  let  U  be  a  square  whose  side  is  the 
linear  unit. 

We  are  to  prove      the  area  of  R  =  a  X  6. 

£-£**  §315 

(two  rectangles  are  to  each  other  as  the  product  of  their  bases  wad  altitudes). 


R 

J>ut  _  is  the  area  of  R, 

. ' .  the  area  of  R  =  a  X  b. 


§316 


Q.  E.  D. 


320.  Scholium.  When  the  base  and  altitude  are  exactly 
divisible  by  the  linear  unit,  this  proposition  is  rendered  evident 
by  dividing  the  figure  into  squares,  each  equal  to  the  unit  of 


T ; 

i             j 



- *• 

-j- 

i 

: 
! 
4— 

i— • 

i 

measure.  Thus,  if  the  base  contain  seven  linear  units,  and  the 
altitude  four,  the  figure  may  be  divided  into  twenty-eight 
squares,  each  equal  to  the  unit  of  measure;  and  the  area  of 
the  figure  equals  7X4. 


COMPARISON     AND    MEASUREMENT    OF    POLYGONS.         179 

Proposition  IV.     Theorem. 

32  1 .    T/n-  area  of  a  parallelogram  is  equal  to  the  product 
of  its  base  and  attitude. 

BE  C     F        B  C    E  F 


A  D 

Let  A  E  FD  be  a  parallelogram,  A  D  its  base,  and  CD 
its  altitude. 

We  are  to  prove      the  area  of  the  O  A  E  F D  =  A  D  X  C  D. 

Prom  A  draw  A  B  II  to  DC  to  meet  F E  produced. 

'Hun  the  figure  A  B  C  D  will  be  a  rectangle,  with  the  same 
base  and  altitude  u  the  CD  A  E  FD. 

In  thert.  A  ABE  and  OPFf 

AB  =  CD,  §126 

ig  opposite  sides  of  a  rectangle). 

and  A  B  =  D  F,  §  134 

(being  opposite  sides  of  a  CD)  \ 

.'.AA/JE  =  A  CDF,  §  109 

/•/.  &  are  equal,  when  the  hypotenuse  and  a  side  of  the  one  are  equal 
reap*  'he  hypotenuse  and  a  side  of  the  other). 

Take  away  the  A  C D F and  we  have  left  the  rect.  ABC D. 

Take  away  tin*  A  A  B  E  and  we  have  left  the  O  A  EFD. 

.'.  ivct.  ABC  D  =  0  A  EFD.  Ax.  3 

Bui  the  ana  of  the  rect.  ABCD  =  ADXCD,     §  319 
{the  area  of  a  rectangle  equals  the  product  of  its  base  and  altitude). 

.'.  the  area  of  the  O  A  E FD  =  A  D  X  C  D.         Ax.  1 

Q.  E.  D. 

'.   Corollary  1.    Parallelograms  having  equal  bases  and 
equal  altitudes  are  equivalent. 

323.     Cor.   2.    Parallelograms    having    equal   bases   are   to 
each  other  as  their  altitudes  ;  parallelograms  having  equal  alti- 
tudes arc  to  each  other  as  their  bases;  and  any  two  parallelo- 
qs  are  to  each  other  as  the  products  of  their  bases  by  their 
altitudes, 

• 


180  GEOMETRY. BOOK    IV. 


Proposition  V.     Theorem. 

324.  The  area  of  a  triangle  is  equal  to  one-half  of  the 
product  of  its  base  by  its  altitude 


Let  ABC  be   a    triangle,  AB  its    base,    and  CD  its 
altitude. 

We  are  to  prove     the  area  of  the  A  A  B  C  =  J  A  B  X  CD. 
From  Cd  raw  OH  II  to  A  B. 
From  A  draw  A  II  II  to  B  C. 

The  ligure  A  B  C  II  is  a  parallelogram,  §  136 

(having  iii  opposite  sides  parallel) •> 

and  A  C  is  its  diagonal. 

.-.  A  A  nC  =  A  A  IIC,  §  133 

{the  diagonal  of  a  CD  divides  it  into  two  equal  &  ). 

The  area  of  the  CJ  A  B  C II  is  equal  to  the  product  of  its 
base  by  its  altitude;  §  321 

.'.the  area  of  one-half  the  £7,  or  the  A  A  B  C,  is  equal  to 
one-half  the  product  of  its  base  by  its  altitude, 

or,  hABX  CD. 

Q.  E.  D. 

325.  Corollary  1.  Triangles  having  equal  bases  and  equal 
altitudes  are  equivalent. 

326.  Cor.  2.  Triangles  having  equal  bases  are  to  each  other 
noir  altitudes  ;  triangles  having  equal  altitudes  are  to  each 

other  as  their  bases;  any  two  triangles  are  fco  <'a<-li  oilier  as  the 
product  of  their  bases  by  their  altitudes. 


<  oMPARISON    AND    MEASUREMENT    OF    POLYGONS. 


181 


Propositi  (an   VI.     Theorem 


;.  The  an 

9Um  of  the  parallel  tide* 


trapezoid  is  equal  to  one-half  the 
oiled  hi/  the  altitude. 
E 


Let  A  HO  If  be  a  trapezoid,  and  E  F  the  altitude. 
H     on  tf prove      area  of  ABC  11=  A  (I1C  +  A  B)  E  l\ 
Draw  the  diagonal  A  t '. 
of  the  A  A  II  ('  =  J  II C  X  E  F,      §  324 

Tea  of  a  A  '  of  Us  boat  by  its  altitude), 

ad  the  area  i  fthe  A  A  BO      .1  .1  BX  A7/7,        §  324 
,  AHC+A  A  IIC, 
otA  BOH      .'.<//('  r  A  B)  /:/'. 

Q.  E    D. 

\i;v.    Th  :  ;i  trapezoid  La  equal  to  the 

product  of  the  line  joining  the  middle  points  <>f  the  non-parallel 

-  multiplied  by  the  altitude;  for  the  line  OP,  joining  the 

middle   points  of   the   non  parallel  sides,   ia  equal   to   h  (II C 

1  142 
,\  by  substituting  0  P  For  \(HC  u  .1  B),  we  have, 
the  area  of  .1  BC 11=  OP  X  /:/■'. 

ffOLIUM.    Tlif 

I  may  be 

iding  the  polygon 
into  triangles,  and  by  finding 
the  area  of  each  of  these  tri- 
angl  >  ly,       But    the 

method  generally  employed  in 
practice  ia  to  draw  the  l<>n. 

diagonal,  and  to  l»-t   fall  perpendiculars  upon  this  diagonal  from 
ilar  points  of  the  polygon. 

The  polygon  is  thus  divided  into  figures  which  are  right 
triangles,  rectangles,  or  trapezoids  ;  and  the  areas  of  each  of  these 

3  may  be  readily  found. 


182  GEOMETRY. BOOK   IV. 


Proposition  VII.     Theorem. 

330.  The  area  of  a  circumscribed  polygon  is  equal  to  one- 
half  the  product  of  the  peri 'meter  by  the  radius  of  the  in- 
scribed circle. 

B 

J) 


Let  ABSQ,  etc.,  be  a   circumscribed  polygon,  and  C 
the  centre  of  the  inscribed  circle. 

Denote  the  perimeter  of  the  polygon  by  P,  and  the  radius 
of  the  inscribed  circle  by  R. 

We  are  to  prove 

ihe  area  of  the  circumscribed  polygon  =  I  P  X  Rt 

Ih-aw  C  A,  CB,  C S,  etc.; 

also  draw  CO,  CD,  etc.,  J_  to  A  B,  B S,  etc* 

The  area  of  the  A  C  A  B  =  \  A  B  X  C  0,  §  324 

(the  area  of  a  A  is  equal  to  oiu-lmlf  the  product  of  its  base  and  altitude). 

The  area  of  the  A  CBS=),BSX  CD,  §  324 

.*.  the  area  of  the  sum  of  all  the  A  CAB,  CBS,  etc., 
=  i  (AB  +  BS,  etc.)  CO,  §  187 

(for  CO,  CD,  etc.,  are  equal,  being  radii  of  the  tame  O). 

Substitute  for  AB  +  BS+SQ,  etc.,  P,  and  for  CO,  R; 

then  the  area  of  the  circumscribed  polygon  =  I  P  X  R. 

Q.  E.  D. 


<<>MPARIS0N    AND    MEASUREMENT    OF    POLYGONS. 


183 


Proposition  VIII.     Theorem. 

881.  The  mm  of  the  squares  described  on  the  two  sides 
of  a  right  triangle  is  equivalent  to  the  square  described  on  the 
hypotenuse. 


Let  ABC  be  a  right  triangle  with  its  right  angle  at  C. 
We  am  to  pro*      A  c'2  +  CB2  =  AB'2 
Draw  00 ±  to  A  />'. 

Then  Tf  =  A  0  X  A  /;.  §  289 

(the  s<i  t  A  is  equal  to  the  produei  of  tic-  hypotenuse  by 

In  i  the  J_  It  full  from  the  vertex  of  the  rt.  Z.)  ; 

and  1TC'2=  BO  X  A  B,  §289 

By  adding,  Uf  +  JUJ2=  (A  0  +  BO)  AB, 
=  ABX  AB, 

=  .  O". 

Q.  E.  D. 

obollabt.   The  side  and  diagonal 
incommensurable. 

Let   A  B  OB   be  a  square,  and  A  C  the 
diagonal. 

Then         A  ir  ■   SJf  =  re2. 

or,  2  fit  —  iT2  5 

Divide  both  sides  of  the  equation  by  ATE2, 

Extract  the  square  root  of  both  sides  the  equation, 

then  ^  =  V  T 

Since  the  square  root  of  2  is  a  number  which  cannot  be 
■My  found,  it  follows  that  the  diagonal  and  side  of  a  square 
wo  incommensurable  Lin 


184 


GEOMETRY.  - 


-BOOK    IY. 


Another  Demonstration. 

883.   The  square  described  <>n   f/ie  hypotenuse  of  a  right 
triangle  is  equivalent  to  the  sum  of  the  squares  on  /he  of  her 

{WO  Sides. 


D  i.    i: 

Let  A  BO  be  a  right  A,  having  the  right  angle  B  A  C. 

Wt  (ire  to  prove      Elf  =  B  A*  +  A  if. 
On  BC>  C A,  A  B  construct  the  Bquarea  B  B,  Off,  A  F. 
Through  A  draw  A  I  I!  to  0  A'. 

Draw  A  1)  and  FC. 

£BAC  ia  a  ri.  Z,  Hyp. 

and  Z  B A  G  ia  a  rt.  Z,  Cons. 


Also 


Now 


,\  C A  '  \ aighl  line 

Z  GA  II  Is  a  it.  Z, 

■.  II  A  II  is  a  straight  Una 

Z  I)BC  =  Z  /<7/,\l, 
rt  /» ///7  a  /•/.  Z). 


(  '(HIS. 


(  Ions. 


COMPARISON    AND    MEASUREMENT    OF    POLYGONS. 


185 


§  106 


Add  to  each  the  Z  A  B  C ; 
then  Z  ABD  =  /.FBC, 

.'.AABD  =  AFBC. 
O  B  Lis  double  A  A  B  D, 
fig  on  tlie  same  base  B D,  and  between  the  same  lis,  A  L  and  B D)> 

and  square  A  F  is  double  A  F B  C, 
* 'j  on  the  same  base  FB,  and  between  the  same  lis,  FB  and  G  C) ; 

.-.  O  B  L  =  square  A  F. 

In  like  manner,  by  joining  A  E  and  B K,  it  may  be  proved 

that 

O  C£  =  square  C II. 

Now  the  square  on  BC^OBL  +  OCL, 

=  square  A  F  +  square  C  H, 


=  BA*  +  AC2. 


Q.  E.  D. 


mono*, 

334.    I>ii.    Tki  P  :>>t  upon  a  straight  line 

of  indefinite  length  is  the  foot  of  the  perpendicular  let  fall  from 
the  point  upon  the  line.  Thus,  the  projection  of  the  point  C 
upon  the  lii)''  A  B  is  the  point  P. 

c  c 


A—p-  — £—     -b       "—jr  —jy 

Fig.  2. 

>/i  of  a  Finite  Straight  Line,  as  C  D  (Fig.  1), 
upon  a  straight  line  of  indefinite  length,  as  A  B,  is  the  part  of 
the  lin«'  A  /;  intercepted  between  the  perpendiculars  C F  and 
I)  L\  lei  tall  from  the  extremities  of  the  line  CD. 

Thus  tin-  projection  of  the  line  C  D  upon  the  line  A  B  is 
the  line  P  //. 

If  one  extremity  of  the  line  CD  (Fig.  2)  be  in  the  line 
A  />'.  the  projection  of  the  line  CD  upon  the  line  A  B  is  the 
part  of  the  line  .1  B  between  the  point  D  and  the  foot  of  the 

radicular  C  P  ;  that  is,  D  P. 


186 


GEOMETRY. BOOK   IV. 


Proposition  IX.     Theorem. 

335.  In  any  triangle,  the  square  on  the  side  opposite  an 
acute  angle  is  equivalent  to  the  sum  of  the  squares  of  the  other 
two  sides  diminished  by  twice  the  product  of  one  of  those 
sides  and  the  projection  of  the  other  upon  that  side. 


Let  C  be  an   acute  angle   of  the   triangle  ABC,  and 
DC  the  projection  of  AC  upon  BC. 

We  are  to  prove      AB2  =  EC2  +  AC*  —  2  B  C  X  D  C. 
If  Z>  fall  upon  the  base  (Kg.  1), 

DB  =  BC-  DC; 
If/)  fall  upon  the  base  prodim-d  (Fig,  2), 

DB  =  DC-BC. 

In  either  case  DH2  =  FT?  +  DTf  -  2  B C  X  I)  C. 
Add  AD   to  both  sides  of  the  equality  ; 

then,  AT/  +  Dl?  =  TfC1  +  JTD2  +  ITD2  -2BCX  DC. 


Lut 


AD2  +  DB2  =AB2 


§331 


{the  sun)   of  tlie  squares  on  two  sides  of  a  rt.  A  is -equivalent  to  the  square 
mi  the  hypotsnn 


and 


AD2  +  ire2  =  ATC2, 


§  331 

Substitute  ATB   and  A  C    for  their  equivalents  in  the  above 
equality  j 


then,      il  //  =  /^  C2  +  AC2  -2BCX  1)  <!. 


Q.  E.  D. 


COMPARISON    AND    MEASUREMENT    OF    POLYGONS.         187 


Proposition  X.     Theorem. 

336.    In    '/////   obtuse   triangle^   the  square  on  the  side 

opposite   the   obtuse  angl*  valent  to  the  sum   of  the 

/'  tlw  oilier  two  sides  increased  hi/  twice  the  product 

of  one  of  those  sides  and  the  projection  of  the  other  on  that 

A 


IT  -* 

Let  C  be  the  obtuse  angle  of  the  triangle  ABC,  and 
CD  be  the  projection  of  A  0  upon  BC  produced. 

We  are  to  prove      JTB2  =  BC2  +  AC1  +  2  B  C  X  D  C. 

D  /i  =  BC  +  DC 

Squaring,  TTB*  =  BC1  +  UT?  +  2BC  X  DC. 

both  sides  of  the  equality; 

then,  A'lr  +  fl  /?  =  BC2  +  A~D2  +  DC2  +  2BCXDC. 

Bat  .1  //"'  +  UB2  =ATB2,  §331 

of  ttu  squares  on  two  sides  of  a  rt.  A  is  equivalent  to  the  square 
on  thr  hypotenuse); 

and  JTB2  +  DC2  -  AC2.  §  331 

Substitute    A  If   and    J~C     for   their    equivalents    in   the 
!  mil  it  v  : 

then.      i  7/2  =  BC2  +  .IC2  +  2  BC  X  D  C. 

Q.  E.  D. 

337.  I>i:i  imthin.  A  Medial  line  of  a  triangle  is  a  straight 
line  drawn  from  any  vertex  of  the  triangle  to  the  middle  point 
of  I  :  le. 


188  GEOMETRY. BOOK    IV. 


Proposition  XI.     Theorem. 

338.  In  any  in 'angle,  if  a  medial  line  be  drawn  from 
the  vertex  to  the  base  : 

I.  The  sum  of  the  squares  on  the  two  sides  is  equivalent 
to  twice  the  square  on  half  the  base,  increased  by  twice  the 
square  on  the  medial  line ; 

IT.  The  difference  of  the  squares  on  the  two  sides  is 
equivalent  to  twice  the  product  of  the  base  by  the  project ion 
of  the  medial  Hue  upon  the  base, 

A 


In  the  triangle  A  EC  let  A  M  be  the  medial  line  and 
M  J)  the  projection  of  A  M  upon  the  base  B  C. 
Also  let  A  B  be  greater  than  A  C. 

We  are  to  ]>rove 

I.    AT?  +  AC1  =  2  1UI2  +  2  A~JP. 

1 1.    IB*  -  Uf  =  2  BOX  M  />. 

Since   A  B  >  A  (',   the  Z  A  MB   will   be   obtuse   and  the 
Z  A  M  C  will  be  acute. 

Then      A  /f      /;  j/-  +  AT)/'1  +  2  BM  X  Mb,       §  336 

my  obtuse  A  tht  square  <>n  tht  tide  <>/>/»>*(/>■  ///'•  obtusi  Z.  is  equivalent  to 
tin  sum  of  tht  >  thr  other  two  Bides  increased  by  twice  the 

product  of  one  of  tho*  <  the  projection  of  the  other  on  that  side)  ; 

and      17?  =  Ml/2  +  UP  -  2  M  C  X  MD,  §  335 

in  any  f .  the  squan  on  th*  side  opposite  an  acute  Z  is  equivalent  to  the  sum 
of  the  squares  on  the  other  two  sides,  diminished  by  tunes' the  product 
of  on>  d  tht  projection  of  the  other  cjimi  flint  side). 

A<bl  these  two  equalities,  and  obs.-rvc  that  //J/  =  M  0. 

Then        Al?  +  AC2  =  2  BM2  +  2  A~M2. 

Subtract  the  second  equality  from  the  first. 

Then  Ali2  -  AC2  =  2BCX  M  I). 

Q.  E.  D. 


COMPARISON    AND    MEASUREMENT    OF    POLYGONS.  189 

Proposition  XII.     Theorem. 
839.   The  sum  of  the  squares  on  the  four  sides  of  any 
quadrilateral  is  equivalent  to  the  sum  of  the  squares  on  the 
dingo, i uls   together  with  four  times  the  square  of  the  line 
joining  (he  middle  point*  of  the  diagonals* 
A 


In  the  quadrilateral  A  Ji  C  D,  let  the  diagonals  be  AC 
and    B  D%    and    VE    the   line  joining   the   middle 
points  of  the  diagonals. 
\Y,  an  to  ft-ove 

rir       /;  ( *2  +  CD2  +  ITj2  —  AC2  +  BT?  +  4  EF1' 
Draw  8  ffand  0  £ 


)'  r   2  HE1, 


338 


{the  sum  of  the  squares  on  Ou  two  sides  twice  the  square 

on  iiaif  th>  has,-  increased  '  to  the  base), 

an.l  +  JJ21  -  2  ( '[''Y  +   2f)E2.  §  338 

A  I  I '.:       ■  '  •  -•  fcwo  <'<(iia]il 

.1  /;J  •    5  (  j  +  i  •  h1  +  0  J'  -  1  (££V  +  2  (jO2  +  £^2). 

BE1  +  hf:1  -  2  (— V  +  2I72,  §  338 

m  equivalent  to  twice  the  square 
on  half  the  base  increased  ;  tc  on  tJie  medial  line  to  the  base). 

Substitute  in  the  aboye  equality  for  (BE    +  BE)  its 
equivalent  ; 

then  AT?  +  JW2  +  TH?  +  ZO2  =  4  (^_)2  +  4  (^)2  +  4  ^2 

=  J~£2  +  BD2  +  4  f?2 

Q.  E.  D. 

3  10.   ( JoBOLLABY.    The  sum  of  the  squares  on  the  four  sides 
of  a  parallelogram  is  equivalent  to  the  sum  of  the  squares  on  the 
aals. 


190 


GEOMETRY. BOOK    IV. 


Proposition  XIII.     Theorem. 

341.  Two  triangles  having  an  angle  of  the  one  equal  to 
an  angle  of  the  other  are  to  each  other  as  the  products  of  the 
sides  including  the  equal  angles. 


Let  the  triangles  ABC  and  A  1)  E  have  the  common 
angle  J. 


W\  <n-e  to  prove 
Now 


A  A  U  ( •        AB^  X  AG 
A  A  J) E  ~  A  1)  X  ARl 


A  ABC 


AC 
AE' 


a  a  n  /■: 

(tk  having  the  same  altitude  arc  to  <<t>-h  other  as  their  bases). 


Also 


A  A  B  /: 


AB 
AD* 


A  A  h  E 
(A  having  the  same  altitude  arc  to  each  other  as  their  bases). 

Multiply  these  equalities ; 

A  ABC  _  ABX  AC 


32G 


32G 


then 


AADE 


AJJXAE 


Q.  E.  D. 


COMPARISON    AND    MEASUREMENT    OF    POLYGONS.        191 

Proposition  XIV.     Theorem. 

342.  8imilar  Mangle*  are  to  each  other  as  the  squares 
on  the'/r  homologous  si<l<.s. 


AL ft XD  A>* £ ±B> 

Let  the  two  triangles  be  AC  B  and  A'C'B'. 

AACB  .1  /r 

We  ore  topr**     -_^_  _. 

Draw  tin*  perpendiculars  CO  and  f'O'. 

A  ACB  A  BX  CO  AB   x    CO_        *  326 

1,11     A  A'C'B'*     A   B'XC'O'        A'B'        CO1'      9 
[two  &  are  to  each  other  as  the  products  of  their  bases  by  their  altitudes). 

AJ*   =    C0-,  §  297 

A'B'        CO''  * 

homologous  altitudes  of  similar  A  have  the  same  ratio  as  their  homolo- 
gous bases). 

Substitute,  in  the  above  equality,  for  -_-  its  equal  -jj-=J 
AACB         AB         AB  ..   Al? 

tll('11         A      „„.„   =    -77-77,    X 


.     .1   C'B'        A'B'  ,s  A'B'        £rp> 

Q.  E.  D. 


192 


GEOMETRY. BOOK    IV. 


Proposition  XV.     Theorem. 

343.   Two   similar  polygons   are   to   each   other  as  the 
squares  on  any  two  homologous  sides. 
B  C 


F  E 

Let    the    two   similar  polygons   be    A  B  C,    etc..    and 
A' BO,  etc. 

w         .                ABC,  etc.          A~l? 
We  are  to  prove =  . 

A'B'C,  etc.        jFjp* 
From  the  homologous  vertices  A  and  A'  draw  diagonals. 

AB         BC         CD 

A'B' 


Now 


etc., 


B'  C        C  ])' 
{similar  polygons  have  (hi  if  homologous  sides  proportional); 


.'.  by  squaring, 


n? 


BC 


CD2 


A~B/2        F  (  C"  D'2 


,  etc. 


The  A  ABC,  A  CD,  etc.,  are  respectively  similar  to  A'B'C, 

A'  C  D',  etc.,  §  294 

(two  similar  polygons  <  Hie.  mine  number  of  &  similar  to  each 

other  and  similarly  placed). 


A  ABC 

A  A'B'C 


AHB'2 


§342 


(similar  &  arc  to  each  other  as  the  squares  on  tlieir  homologous  sides), 
AACD  CTJ2 


and 


A  A'  C  D'    "  tf-tf 


§342 


COMPARISON    AND    MEASUREMENT    OF    POLYGONS.        193 


But 


CI? 

AT? 

unr- 

A^' 

AABC 

A  A'B'C 

A  ACD 

A  A'  CD' 

In  like  manner  we  may  prove  that  the  ratio  of  any  two  of 
the  similar   1\  La  the  same  as  that  of  any  other  two. 

.    _AABC  \  CD  A  APE  AAEF 

A  A'B'C'  ~~  A  A' CD'  ~~  AA'D'E  "  AA'E'F' 

ABC  +  AODMADB+  ABF  A  ABC 


A  A'  B'  C  +  A'  CD'  4-  A'  D'  E'  +  A'  E'  P      A  A' B'  C 

I  aeries  of  equal  ratios  the  sum  of  the  antecedents  is  to  tlie  sum  of  the 
consequents  as  any  antecedent  is  to  its  consequent). 

t>   *  A  ABC  H?  t  Q,0 

Hut  =  ,  §  342 

A  A'B'C        A   jf  * 

U&r  &  are  to  each  other  as  the  squares  on  f/m'r  Jionwlogous  sides) ; 

.     the  polygon  ABC,  etc.       _  A~7? 
the  polygon  A'  B'  C,  etc.        A'  B'2 

Q.  E.  D. 

34  I.    COBOLLABT    1.      Similar  polygons  are  to  each  other  as 

th'-  s.ju  n,  .  mii  anv  two  homologous  lines. 

345,    Cob.  2.    He  homologous  sides  of  two  similar  poly- 
have  the  Bame  ratio  as  the  square  roots  of  their  areas. 

I  9  represent  the  areas  of  the  two  similar  polygons 
A  B  C,  etc.,  an*  1  A'  B'  C,  etc.,  respectively. 

Then  S  :  S'  ::  AB2  :  A7^2, 

hjgons  arc  to  each  other  as  the  squares  of  their  homologous  sides). 

s[S  :  <Jtf  :  :  AB  :  A' B'y  §  268 

or,  AB  :  A' B>  :  :  \[S  :  sfS1. 


194 


GEO^IETRY. BOOK    IV. 


On  Constructions. 

Proposition  XVI.     Problem. 

346.  To  construct  a  square  equivalent  to  the  sum  of  two 
given  squares. 


A     ~fr- 

Let  R  and  R'  be  two  given  squares. 
It  is  required  to  construct  a  square  =  R  +  R'. 
Construct  the  rt.  Z  A. 

Take  A  B  equal  to  a  side  of  R, 

and  A  C  equal  to  a  side  of  R'. 

Draw  B  C. 

Then  B  C  will  he  a  side  of  the  square  required. 

For  ~BC2  =  TB2  +  TC2,  §  331 

{the  square  on  the  /<  of  a  rt.  A  ie  equivalent  to  the  own  of  the 

ires  on  tht  two  Bid 

Construct  the  square  Sf   haying  each    of   its  sides  equal 
to  BC. 

Substitute   for'^6'2,   AB2   and   AC2,  S,   R,   and  R'  re- 
spectively ; 

then  S=  R+  R'. 

.*.  S  is  the  square  required. 

Q.  E.  F. 


CONSTRUCTIONS. 


195 


Proposition  XVII     Problem. 

347.   To  co 1 1  I  square  equivalent  to  the  difference 

of  too  given  square^. 


B 

R 

w 

A 

V--.  x 

i            i 

!  s   ! 

i               i 

Let  R  be  the  smaller  square  and  R'  the  larger. 
It  is  required  to  construct  a  square  =  R'  —  R. 
•  tin-  rt.  . 

Take  A  B  equal  to  a  side  of  R. 

From  B  as  a  centre,  with  a  radius  equal  to  a  side  of  Rft 

desciil"  an  arc  cutting  the  line  A  X  at  C. 

Then  A  C  will  be  a  side  of  the  square  required. 

draw  B  C. 

AB*  +  Alf  =  IW2,  §  331 

mm  of  the  squares  on  the  two  sides  of  a  rt.  A  is  equivalent  to  the  square 
on  the  hypotenuse). 

By  transposing,  AC1  =  BT?  —  Al?. 

Construct  the  square  S,  haying  each  of  its  sides  equal  to  A  C. 

Substitute   for  AC2,  BV\  and  AB1,   S,  R',  and  R  re- 
stively ; 
then  8  =  R'  —  R. 

.*.  S  is  the  square  required. 

Q.  E.  F 


196  GEOMETRY. BOOK   IV. 

Proposition  XVIII.     Problem. 

348.   To  construct  a  square  equivalent  to  the  sum  of  any 
number  of  given  squares. 


P- 
r- 
m- 


H 

,.<-\ 

y     \ 

,'' 

\ 

FS 

\ 

\ 

/ 

\ 

/ 

\ 

#K 

\ 
\ 

1 

\ 

i                   Xv 

s          \        \ 

\   \  \ 

!        ""*-«— ..^ 

\  \  \ 

j 

*■ — »    X  N>  \ 

V 

— r^kn 

Let  m,  ??,  o,  p,  r  be  sides  of  the  given  squares. 
It  jj  >vy >//'/<  d  to  instruct  a  square  =  m2  4~  n2  -f  o2  +  p2  +  r2. 
Take  A  B  =  w. 

Draw  A  C  =  n  ami  _L  to  A  B  at  A. 
Draw  BC 
Draw  CE=o  and  _L  to  B  C  at  C,  and  draw  B E. 
Draw  BF—ptnA  JL  to  B  E  at  A',  and  draw  BF. 
Draw  FII  =  r  and  ±  to  B  F  at  F,  and  draw  B  JL 
The  square  constructed  on  B II  is  the  square  required. 

For      BH2  =  FTP  +  572, 

—  /^T?'2  +  FT'2  +  yO2, 

=  F712  +  jP72  +  #X'2  +  Cl?, 

=  FTl2  4-  EF2  +  ^6t2  +  6TI2  +  Iff,  §  33 1 

i///'   .<wra  o/  //^  squares  on  two  sides  of  a  rt.  A  is  equivalent  to  ///'•  tquart 
on  the  hypotenuse). 

Substitute  for  A  B,  G  A,  EC,  E F,  and  FII,  m,  n,  o,  pt 

and  r  respectively ; 

then  BTl2  =  m2  +  n2  +  o2  +  />2  +  r2 

Q.  E.  F. 


\  STRUCTIONS. 


197 


Proposition  XIX.     Problem. 

349.   To  construct  a  polygon  similar  to  two  given  similaf 
polygons  and  ,rf  to  their  9UM* 


s. 


\ 


\ 


R» 


A"  13" 


)o-. 


H 

Let  R  and  I!'  be  two   similar  polygons,  and  AB  and 
IV  two  homologous  sides. 

It  /   polygon  equivalent  to 

R  +  R . 

;  tin*  rt.  Z  V. 

Take  /'//  =  A' JV.  rod  ro  =  AB. 
Draw  Off. 

ke  A"  B"  =  0  II. 
Upon  .1    A'  .  homologous  to  jii?,  construct  the  polygon  7?" 
similar  to  /.'. 

Thru  B"  ia  the  polygon  required. 

«    :   //        J^g72  :  .O2,  §  343 

far  polygons  arc  to  each  ollur  as  the  squares  on  their  homologous  sides). 

An*'2. 


Also  S      :  A     :   :    .1     W 

Iii  the  tirst  proportion,  by  composition, 


But 


fi     r  /,    :    W 


R"  :  R' 


FTl1  +  i^2  :  PIT2, 
I/O2  :  P772. 


§  343 
§  264 


HO2  :  P772. 


P"  :  R'  ::  R'  +  R  :  R'; 

.'.  R"  =  R'  +  R. 


Q.  E.  F. 


198 


GEOMETRY. BOOK  IV. 


Proposition  XX.     Problem. 

350.   To  construct  a  polygon  similar  to  two  given  similar 
polygons  and  equivalent  to  their  difference. 


x 

/            ^       Hi- 

/    i   \ 

V       J     1 

A"         B»       P 

.X. 
0 

A'  Bf  A  B 

Let  R  and  R'  be  two  similar  polygons,  and  AB  and 
A' B'   two  homologous  sides. 

It  is  required  to  construct  a   similar  polygon  which  shall 
be  equivali  kt  to  II'  —  R. 

Construct  the  rt.  Z  P, 

and  takeP0  =  A  11 

From  0  as  a  centre,  with  a  radius  equal  to  A'  B', 

describe  an  arc  cut  tin-  1}  X  at  HI 

Draw  0  II. 

Take  A"  B"  =  P  II. 

On  A"  B",  homologous  to  A  B,  construct  the  polygon  R" 
similar  to  R. 

Then  I! '  is  the  polygon  required. 

Rf  :  R  :  :  A'HS'2  :  .O2,  §  343 

(similar  polygons  are  to  each  other  as  the  squares  on  their  homologous  sides). 

Also  R"  :   R   :  :  .FT57'2   :  AT?. 

In  the  first  proportion,  by  division, 


R'-  R  :  R 


:  A'B'2-  AB2  :  IB2, 
:  Oil2  -  OP1  :  07*, 


§343 
§  2G5 


But 


:  PTl2  :  07*. 
R"  :  R  :  :  i77^2  :  AB2, 

:  :  i7/?2   :  OP2. 
.'.  R"   \  R   :  :  R'  -  I!   .    R  ; 


Q.  E.  F. 


I  <>\>TUUCTIONS. 


199 


Proposition  XXI.     Problem. 

351.    To   construct  a   triangle  equivalent   to    a  given 
polygon.  c  D 


I  A  i  r  £■ 

Let  ABC D1I E  be  the  given  polygon. 

'id  to  the  given 
gon. 

m  1)  draw  D  Bt  and  from  //  draw  II F  II  to  DK 

Produce   A  E  to  mfeet  //  /'  at  7^,  and  draw  D  F. 

The  polygon  A  BC  D  F  has  one  side  less  than  the  polygon 
A  BC D  II  E.  l»ut  the  two  are  equivalent. 

For  the  part  A  BCD E  is  common, 

and  the  ADEF  =  ADEII,foY  the  base  D  E is  common, 
and  their  vertdcefl  /'and  //are  in  the  line  /\#  II  to  the  base,     §325 
(A  having  the  same  base  and  equal  altitudes  are  equivalent). 

Again,  draw  OF,  and  draw  D  K  II   to  C  F  to  meet  A  F 
produced  at  K. 

Draw  CK. 

The  polygon  ABC K  has  one  side  less  than  the  polygon 
A  B  C  D  F,  but  the  two  are  equivalent. 

For  the  part  A  B  C  F  is  common, 

and  the  A  C  F K  =  A  CFD,  for  the  base  CF  is  common, 
and  their  vertices  JTand  D  are  in  the  line  KD  II  to  the  base.      §  325 

In  like  manner  we  may  continue  to  reduce  the  number  of 
sides  of  the  polygon  until  we  obtain  the  A  C I K. 

Q.  E.  F. 


200  GEOMETRY. BOOK    IV. 


Proposition  XXII.     Problem. 

352.   To  construct  a  square  which  shall  have  a  given 
ratio  to  a  given  square. 

s''     /Iv\ 

/  /         \       \       \ 

/     /        i      \    s 

m !£/- j. \f\ 


^ -  B iC 

n 
Let  R  be  the  given  square,  and  -   the  given  ratio, 

m 

It  is  required  to  construct  a  square  which  shall  be  to  R  as 

n  is  to  77i. 

On  a  straight  line  take  AB  =  m,  and  B  C  =  n. 

On  AC  as  a  diameter,  describe  a  semicircle. 

At  B  erect  the  ±  B  S,  and  draw  8  A  and  SC. 

Then  the  A  A  SC  is  a  rt.  A  with  the  it.  Z  at  S,     §  204 
\  iii  a  semicircle.) 

On  SA,  or  SA  produced,  take  SE  equal  to  a  side  of/?. 

Draw  i?/7  II  to  AC. 

Then  -s'  /T  La  a  aide  of  the  square  required. 

For  £?!  =  :!*  §289 

SC*       "C  * 

(the  squares  on  tJie  sides  of  a  rt.  A  have  tin  tame  ratio  as  tJie  segments  of  the 
hypotenuse  wade  by  the  _L  let  fall  from  the  t*  Hex  of  the  rt.  Z). 

Also  SA  =  -S  A  .  §  275 

(a  straight  line  drawn  through  two  sides  of  a  A,  parallel  to  the  third  side, 
divides  those  tides  proportionally). 

Square  the  last  equality  ; 

then  **-g. 

SI?      M2 

q—r2  g-rp 

Substitute,  in  the  first  equality,  for  -J—^  ^s  eclual  -    \,  > 


then 


Si?  ST 

SI?         AB        m 


S  /.--•  BC         n 

that  is,  the  square  having  a  aide  equal  to  8JFvri31  have  the 
same  ratio  to  the  square  Ef  as  n  has  to  m. 


Q.  E.  F. 


CONSTRUCTIONS.  201 


Proposition  XXIII.     Problem. 

353.  To  construct  a  polygon  similar  to  a  given  polygon 
and  having  a  .  itio  to  it. 


!\\      /  \ 

: -\F\       (        s       > 

i ^     \  / 

m \^_  J 

n A'   ~       B' 

Let  R  be  the  given  polygon  and  -   the  given  ratio. 

It   is  required  to  construct  a  polygon  similar  to  R,  which 
sh<i/i  in  h,  i;  eu  n  if  to 

Find  a  line,  A   A',  >inh  tliat  the  square  constructed  upon  it 
shall  1m-  t«»  the  square  constructed  upon  A  B  as  n  is  to  m.     §  352 

Upon  A'  B'  as  a  side  homologous  to  A  B,  construct  the 
•  >n  s  similar  to  R. 

Th. -11  8  is  the  polygon  required. 

For  f-£g,  §343 

«         A  J? 
liar  polygons  are  to  each  other  as  tlie  squares  on  their  homologous  sides). 

But  £*  .  *  5  Cons. 

Al?        m 

.'.-  =  -,     or,     S  :  R  :  :  n  :  m. 

R  7)1  \ 

Q.  E.  F. 


202  GEOMETRY. BOOK   IV. 


Proposition  XXIV.     Problem. 

354.  To  construct  a  square  equivalent  to  a  given  paral- 
lelogram. 

P 

b  c  r 1  --T \ 

•■I  /!     \ 

-I     m  L l * V 

M  N  0     X 

Let  ABC  I)  be  a  parallelogram,  b  its  base,  and  a  its 
altitude. 

It  is  required  to  construct  a  square  =  O  ABC D. 

Upon  the  line  MX  take  MN  =  a,  and  N  0  m  b. 

Upon  M  0  as  a  diameter,  describe  a  semicircle. 

At  iVr  erect  i\TPJ_  to  MO. 

Then  the  square  ft,  constructed  upon  a  line  equal  to  N  P, 
is  equivalent  to  the  O  A  BCD. 

For  MN  :  NP  :  :  NP  :  NO,  §  307 

{a  ±  let  fall  from  any  poind  iff  a  circumference  to  Oie  diameter  is  a  mean 
proportional  between  the  segments  of  the  diameter). 

r.NI^^MNX  NO  =  aXb,  §  259 

(the  product  of  the  means  is  equal  to  the  product  of  the  extremes). 

Q.  E.  F. 

355.  Corollary  1.  A  square  may  be  constructed  equiva- 
lent to  xa  triangle,  by  taking  for  its  side  a  mean  proportional 
between  the  base  and  one-half  the;  altitude  of  the  triangle. 

356.  Cor.  2.  A  square  may  be  constructed  equivalent  to 
any  polygon,  by  first  reducing  the  polygon  to  an  equivalent  tri- 
angle, and  tlien  constructing  a  square  equivalent  to  the  trian 


CONSTRUCTIONS. 


203 


Proposition  XXV.     Problem. 

867.  To  construct  a  parallelogram  equivalent  to  a  given 
\re%  and  having  tie  sum  of  its  base  and  all  Hade  equal  to 
ven  Hn<'. 


M 


/ 


& 


Let  fi  be  the  given  square,  and  let  the  sum  of  the 
base  and  altitude  of  the  required  parallelogram 
be  equal  to  the  given  line  M  N. 

It  i  a  UJ  =  //,  and  having  the  sum 

El  base  and  z  M  X. 

Upon  M  N  as  a  diameter,  describe  a  semicircle. 
.!/  erect  a  _L  M  P,  equal  to  a  side  of  the  given  square  JR. 
Draw  PQ  II  to  MX,  cutting  the  circumference  at  S. 

DrawSC-Lto  J/^. 
Any  O  having  CM  tot  its  altitude  and  CN  for  its  base, 
is  equivalent  to  R. 

SC  is  II  to  P M,  §-65 


{two  straight  lines  JL  to  the  same  straight  line  are  II ). 

..sc  =  pm, 

(lis  comprehended  between  lis  are  equal). 


§135 


Bat  M  0  :  SC  :  :  SC  :  C  N,  §307 

(a  J_  let  fall  from  any  point  in  a  el.  ie$  to  the  diameter  is  a  mean 

proportional  between  the  segments  of  the  diameter). 

Then  8Ci  =  MCXCN,  §259 

{the  product  of  the  means  is  equal  to  the  product  of  the  extremes). 

Q.  E.  F. 

358.  Scholium.    The  problem  is  impossible  when  the  side 
of  the  square  Lb  greater  than  one-half  the  line  M  N. 


204 


GEOMETRY. BOOK  IV. 


Proposition  XXYI.     Problem. 

359.  To  construct  a  parallelogram,  equivalent  to  a  given 
square,  and  having   the  difference  of  its  base  and  altitude' 
equal  to  a  given  line, 
S 


\c . 


M\ 


-?JV 


/ 
/ 

T 

/ 

/ 
/ 

/ 

Rf 

/ 

'-  "D 
Let  R  be  the  given  square,  and  let  the  difference  of 
the  base  and  altitude  of  the  required  parallelo- 
gram be  equal  to  the  given  line  M AT. 

It  is  required  to  construct  a  O  =  //.   with  the  difference 
of  the  base  and  altitude  =  M  N. 

Upon  the  given  line  M N  as  a  diameter,  describe  a  circle. 

From  M  draw  MS,  tangent  to  the  O,  and  equal  to  a  side 
of  the  given  square  R, 

Through  the  centre  of  the  O,  draw  SB  intersecting  the 
circumference  at  C  and  B. 

Then  any  O,  as  R',  having  SB  tot  its  base  and  SC  for 
its  altitude,  is  equivalent  to  R. 

For  SB  :  SM  :  :  SM  :  SC,  §292 

<>7n  a  point  vnthout  a  O,  a  secant  and  a  tangent  be  drawn,  the  tangent  is 
a  mean  proportional  between  the  whole  secant  and  the  part  without  the  O). 

Then  STl2  =  8  B  X  8  C;  §  259 

and   the    difference  between  S 11  and   80  IB  the  diameter 
of  the  O,  that  is,  M  N. 

Q.  E.  F. 


CONSTRUCTIONS.  205 


Proposition  XXVII.     Problem. 


360.    G  > '  =  sl'l,  to  construct  x. 

\ 


E  .... 


. 


J 


A — t - — D'B 


Let  m  represent  the  unit  of  length. 

It  is  required  to  i  •>  which  shall  represent  the  square 

root  of  2. 

On  the  indefinite  line  A  B%  take  A  C  =  m,  and  CD  =  2  m. 

On  4  Z)  as  a  diameter  describe  a  semi-circumference. 

At  C  erect  a  _L  to  j4  B,  intersecting  the  circumference  at  E. 

Then  C  E  is  the  line  required. 

For  AC  :  CE  ::  CE  :  CD,  §  307 

(£A*  ±  let  fall  from  any  poi  ferenee  to  the  diameter,  is  a  mean 

proportional  between  the  segments  of  tlie  diameter)  ; 

.\C~E2  =  ACXCD,  §259 


CE=sjACX  CD, 


=  >J\  X  2  =  \/2. 

Q.  E.  F. 


Ex.   1.    Given  x  =  )/d,  p  =  sJ7 ,  z  =  2  \/5  ;  to  construct  x,  y, 
and   f. 

2.  Given  2   :  .r  :  :  x  :  3;  to  construct  x. 

3.  Construct  a  square  equivalent  to  a  given  hexagon. 


206  GEOMETRY. BOOK   IV. 


Proposition  XXYIII.     Problem. 

361.  To  construct  a  polygon  similar  to  a  given  polygon 
P,  and  equivalent  to  a  given  polygon  Q. 


\_/ 

At  B' 


m- 


.1 


m       A  B 

Let  P  and  Q  be  two  given  polygons,  and  A  B  a  side 
of  polygon  J\ 

It  is  required  to  construct  a  polygon  similar  to  P  and  equiva- 
lent to  Q. 

Find  a  square  equivalent  to  P,  §  356 

and  let  m  be  equal  to  one  of  its  sides. 

Find  a  square  equivalent  to  Q,  §  356 

and  let  n  be  equal  to  one  of  its  sides. 

Find  a  fourth  proportional  to  m,  n,  and  A  B.  §  304 

Let  this  fourth  proportional  be  A'  B'. 

Upon  A'  B'y  homologous  to  A  B,  construct  the  polygon  P' 
similar  to  the  given  polygon  P. 

Then  P1  is  the  polygon  required. 


CONSTRUCTIONS. 

207 

For 

m         AB 

n   ="  A'  Br 

Cons. 

Squaring, 

m*         AB* 

Bat 

P  =  m2, 

Cons. 

and 

<?  =  *2; 

Cons. 

.  p 

w2       HP 

'•« 

■?      47iT'2 

But  J  -    J  A,  p  §  343 

v  arc  to  c<tcA  other  as  the  squares  on  their  homologous  sides)  / 

..-  =  —:  Ax.  1 

/.  /'  is  equivalent  to  Q,  rimilar  to  P  by  construction. 

Q.  E.  F. 


Ex,    1.    Construe!    a  square  equivalent  to  the  sum  of  three 
giv<!  a  whose  sides  are  respectively  2,  3,  and  5. 

-^2.    Construct   a   square  equivalent  to   the  difference  of  two 
given  whose  sides  are  respectively  7  and  3. 

rtaruci  a  square  equivalent  to  the  sum  of  a  given  tri- 

angle  and  a  given  parallelogi 

_r     4.    Construct  a  rectangle  having  the  difference  of  its  base  and 

en  line,  and  its  area  equivalent  to  the  sum 

of  a  given  triangle  and  a  given  pentagon. 

l      5.    Given  a  hexagon ;  to  construct  a  similar  hexagon  whose 

til  be  to  that  of  the  given  hexagon  as  3  to  2. 

'».    Construct   a   pentagon  similar  to  a  given  pentagon  and 

equivalent  to  a  given  trapezoid. 


208  GEOMETRY. BOOK   IV. 


Proposition  XXIX.     Problem. 

362.   To  construct  a  polygon  similar  to  a  given  polygon, 
and  having  two  and  a  half  times  its  area. 

Y 

A 

// 
// 
/  / 

/ 


M 


LD 


I 
I 


B  C  0  N 

Let  P  be  the  given  polygon. 

It   is   required   to   construct   a  polygon   similar   to  P,  and 
equivalent  to  2£  P. 

Let  A  B  be  a  side  of  the  given  polygon  P. 
Then  ft  :  v^  :  :  A  B  :  x, 

or  <J2   :  )JT)   :  :  A  B  :  x,  §  345 

(the  homologous  sides  of  similar  polygons  are  to  each  other  as  the  square  roots 
of  tlicir  areas). 

Take  any  convenient  unit  of  length,  as  M  C,  and  apply  it 
six  times  to  the  indefinite  line  M  N. 

On  MO  (=  3  M  G)  describe  a  semi-circumference ; 

k 
and  on  M ' N  (=  6  M  G)  describe  a  semi-circumference. 

At  G  erect  a  _L  to  M  iV,  intersecting  the  semi-circumfer- 
ences at  D  and  H. 

Then  CD  is  the  ^2,  and  GH  is  the  yfi.  §  3G0 

Draw  C  F,  making  any  convenient  Z.  with  G  H. 
On  C  Y  take  GE=  A  B. 
From  D  draw  D  E, 
andfrmn  //  draw  H  V  II  to  D  E. 


CONSTRUCTIONS.  209 


Then  C  Y  will  equal  x,  and  be  a  side  of  the  polygon  re- 
quired, homologous  to  A  B. 

For  CD  :  C II  :  :  CE  :  C  Y,  §275 

(a  Kim  cfrawn  through  two  sides  of  a  A,  II  to  tffo  2/urd  mfe,  divides  the  two 
sides  proportionally). 

Substitute  their  equivalents  for  C  D,  C II,  and  C  E ; 

then  )/2   :  fi  :  :  A  B  :  C  Y. 

On  C  Y,  homologous  to  A  B,  construct  a  polygon  similar 
to  the  given  p<»lygon  P; 

and  tins  is  the  polygon  required. 

Q.  E.  F. 


C 


1.  The  perpendicular  distance  between  two  parallels  is 
30,  and  a  line  is  drawn  across  them  at  an  angle  of  45° ;  what  is 
its  length  between  the  parallels  1 

-U2.  m  equilateral  triangle  each  of  whose  sides  is  20; 

find  the  altitude  of  the  triangle,  and  its  area. 

3.  Given  the  angle  A  of  a  triangle  equal  to  §  of  a  right 
angle,  tli  B  equal  to  J  of  a  right  angle,  and  the  side  a, 

opposite  the  angle  As  equal  te  10;  construct  the  triangle. 

-  of  a  chord  intersected  by  another  chord 
are  G  and  5,  and  one  segment  of  the  other  chord  is  3;  what  ~: 
ifl  the  otl  :it  of  the  latter  chord  1 

/   5.    If  a  circlo  be   inscribed  in  a  right  triangle :  show  that 
tlif  difference  between  the  sum  of  the  two  sides  containing  the  A 
right  angle  and  the  hypotenuse  is  equal  to  the  diameter  of  the 
circle. 

6.  Construct  a  parallelogram  the  area  and  perimeter  of  which 
shall  be  respectively  equal  to  the  area  and  perimeter  of  a  given 
triangle. 

7.  Given  the  difference  between  the  diagonal  and  side  of  a 
square;  construct  the  square. 


BOOK  V. 

REGULAR  POLYGONS  AND  CIRCLES. 

363.  Def.    A   Regular  Polygon   is    a    polygon   which    is 
equilateral  and  equiangular. 

Proposition  I.     Theorem. 

364.  /  / ater al  polygon  inscribed  in  a  circle  is  a 

regular  polygon. 

C 

D 


Let  ABC,  etc.,    be  an    equilateral  polygon  inscribed 
in  a  circle. 

We  are  to  prove  the  polygon  ABC,  etc.,  regular. 

The  arcs  A  B,  B  C,  C  D,  etc.,  are  equal,  §  182 

(in  the  same  O,  equal  chords  subtend  equal  arcs). 

.'.  arcs  ABC,  BCD,  etc.,  are  equal,  Ax.  6 

.*.  the  A  A,  B,  C,  etc.,  are  equal, 
{being  inscribed  in  equal  segments). 

.'.  the  polygon   ABC,  etc.,   is  a   regular   polygon,   being 
equflatenl  and  equiangular. 

Q.  E.  D. 


REGULAR   POLYGONS    AND    CIRCLES.  211 


Proposition  II.     Theorem. 

365.  I.    A  circle  may  be  circumscribed  about  a  regular 
polygon. 

II.  A  circle  may  be  inscribed  in  a  regular  polygon. 


B  /^~~' 

r^^%vC' 

Au 

v  ^*^y~_ 

_—^  E 

w 


Let  ABC  D,  etc.,  be  a  regular  polygon. 
We  are  to  prove  that  a  O  may  be  circtcmscribed  about  this 
•on,  and  also  a  O  mag  be  inscribed  in  this  regular 
<jon. 

Case  L  —  Describe  a  circumference  passing  tli rough  A,  B,  and  C. 

From  the  centre  0,  draw  0  A,  0  D, 

and  draw  0  s  J_  to  chord  B  C. 

On  0  s  as  an  axis  revolve  the  quadrilateral  0  A  B  s, 

until  it  cornea  into  the  plane  of  OsC D. 

Tin*  Line  %B  will  fall  upon  sC, 
(for  Z0sB  =  Z0sC,  both  being  rt  A ). 

Tlu-  point  B  will  fell  upon  C,  §  183 

iice  8 B  =  sC). 

He  line  BA  will  fall  upon  CD,  §  363 

Z  B  =  Z  C,  fct'/igr  A  of  a  regular  polygon). 
The  point  4  will  fall  upon  Z>,  §  363 

c  B  A  =  C  D,  being  sides  of  a  regular  polygon). 

.'.  the  line  CM  will  coincide  with  line  0  J), 

{their  extremities  briny  the  same  points). 

.'.  the  circumference  will  pass  through  D. 

In  like  manner  we  may  prove  that  the  circumference,  pass- 
ing through  vertices  B,  C,  and  D  will  also  pass  through  the 
vertex  E,  and  thus  through  all  the  vertices  of  the  polygon  in 
succession. 

B  II.  —  The  sides  of  the  regular  polygon,  being  equal  chords  of 

the  circumscril  >ed  O,  are  equally  distant  from  the  centre,     §  185 

i   circle  described  with  the  centre  0  and  a  radius  Os 

will  touch  all  the  sides,  and  he  inscribed  in  the  polygon.     §  171 


212 


GEOMETRY. BOOK    V. 


366.  Def.    The  Centre  of  a  regular  polygon  is  the  common 
centre  0  of  the  circumscribed  and  inscribed  circles. 

367.  Def.    T\\q  Radius  of  a  regular  polygon  is  the  radius 
0  A  of  the  circumscribed  circle. 

368.  Def.    The  Apotkem  of  a  regular  polygon  is  the  radius 
0  s  of  the  inscribed  circle. 

369.  Def.    The  Angle  at  the  centre  is  the  angle  included 
by  the  radii  drawn  to  the  extremities  of  any  side. 


Proposition  III.     Theorem. 

370.  Each  angle  at  the  centre  of  a  regular  polygon  is 
equal  to  /bur  right  angle*  divided  hjf  the  number  of  sides 
of  the  polygon. 


Let  ABC,  etc.,  be  a  regular  polygon  of  n  sides. 

_  4  rt.  A 
Wi    are  to  prove      Z.  A  0  B  —  — - —  • 

Circumscribe  a  O  about  the  polygon. 

The  A  A  OB,  B  0  C,  etc.,  are  equal,  §  180 

(in  tlie  same  O  equal  arcs  subtend  equal  A  at  the  centre). 

.'.  the  Z  A  0  B  =  4  rt.  A  divided  by  the  number  of  A  about  0. 

But  the  number  of  A  about  0  =  n,  the  number  of  sides 
of  the  polygon. 


A  AOB  = 


4  rt.  A 


Q.  E.  D. 


371.  Corollary.    The   radius   drawn  to  any  vertex  of  a 
regular  polygon  bisects  the  angle  at  that  vertex. 


REGULAR   POLYGONS   AND    CIRCLES.  213 


Proposition  IV.     Theorem. 

2 .   Two  regular  polygons  of  the  same  number  of  sides 
are  similar. 


Q' 


Let  Q  and  Q'  be  two  regular  polygons,   each  having 
n  sides. 

We  are  to  prove       Qa  dar  polygons. 

The    sum   of  the   interior  A  of  each  polygon  is  equal  to 
2  it.  A  (n  -  !  §  157 

(the  m  lor  A  of  a  U  to  2  rt.  A  taken  as  many 

■ygon  luis  sides). 

,     s                                     2  rt.  A  hi—  2)  -1KQ 

Each  Z  of  the  polygon  Q  = -* •  »         §  158 

(for  the  A  of  a  n  en-  all  equal,  and  hence  each  A  is  equal 

•  d  by  their  number). 

AIs ...  each  Z  of  Q'  =  2  rt-  ^  (»  -  2)  §  lg8 

n 

.'.  the  two  polygons  Q  and  Q'  are  mutually  equiangular. 

Moreover,  ^A  =  1>  §  363 

(/Ae  sides  of  a  regular  polygon  are  all  equal) ; 

and  ^=1,  §363 

.    ±B  =  *B_%  Ax.l 

£  C       i?'  C" 

.*.  the  two  polygons  have  their  homologous  sides  proportional ; 
.'.  the  two  polygons  are  similar.  §  278 

Q.  E.  D. 


214 


GEOMETRY. BOOK  V. 


Proposition  V.     Theorem. 

373.  The  homologous  sides  of  similar  regular  polygons 
have  the  same  ratio  as  the  radii  of  their  circumscribed  cir- 
cles, and,  also  as  the  radii  of  their  inscribed  circles. 


Let  0  and  0'  be  the  centres  of  the  two  similar  regu- 
lar polygons  ABC,  etc.,  and  A'B'C,  etc. 

Erom    0  and    O1  draw   0  E,   0  D,   O'E1,   0'  D',    also   the 
Js  0  m   and    0'  m'. 

0  E  and  0'  E'  are  radii  of  the  circumscribed  (D,         §  367 

and  0  in  and  0'  m'  are  radii  of  the  inscribed  ©.         §  3G8 

ED  <>/■:  Om 

=  0>  FJ  - 


We  are  to  prove 


ED'        0>  E'        O'm' 
In  the  A  0  E  1)  and  0'  E'  D' 

the  A  0  ED,  ODE,  O1  E D'  and  0'  D'  E  an  equal,     §  371 

(being  halves  of  the  equal  A  F  E  D,  E 1)  C,  F'  E'  Df  and  Ef  D'  O) ; 

.-.  the  &OE  I)  and  C  #  /)'  are  similar,  §  280 

{if  two  &  have  two  A  of  the  one  eq  ucU  >  to  two  A  of  the  other,  they 

are  similar). 

K  I)  OE 


Also, 


ED'    "  O'E'' 
(the  homologous  sides  of  similar  A  o.re  proportional). 

El)  Om 


E  I)' 


O1  mi 


§278 


§  297 


(the  homologous  altitudes  of  similar  &  have  the  same  ratio  as  tlieir  homolo- 

Q.  E.  D. 


REGULAR   POLYGONS    AND    CIRCLES. 


215 


Proposition  VI.     Theorem. 

374.  The  perimeters  of  similar  regular  polygons  have 
He  name  ratio  as  the  radii  of  their  circumscribed  circles,  and, 
"/■so  as  the  radii  of  tit  (bed  circles, 

s0 


Let  P  and  V  represent    the  perimeters    of  the    two 

similar  regular  polygons  ABC,  etc.,  and  A'B'C,  etc. 

raw  0  E,  0'  E',  and  _l§  0  m  and  0'  ml. 

Om 


m  «  p         OE 

We  are  to  prove      —  =  

1  P>        0>  E' 


p_ 


ED 

E'  D' P 


§  295 


(the  perimeters  o)  polygons  have  the  same  ratio  as  any  two  homolo- 

gous sides). 


Moreover,  = 


OE 

o  /■: 


ED^ 
E'  D1' 


§373 


{the  homologous  sides  of  similar  regular  polygons  have  the  same  ratio  as  the 

rcumscribed  CD). 


Also 


Om 
O'm' 


ED 


§373 


(the  homologous  sides  of  similar  regular  polygons  have  the  same  ratio  as 
the  radii  of  their  inscribed  (D). 


P 
P' 


OE 

We> 


Om 
0^' 


Q.  E.  D 


216 


GEOMETRY. BOOK   V. 


Proposition  VII.     Theorem. 

375.  The  circumferences  of  circles  have  the  same  ratio 
as  their  radii. 


Let  C  and  C  be    the    circumferences,  B  and  E'   the 
radii  of  the  two  circles  Q  and  Q'. 

We  are  to  j>rove      C  :  C"   :  :  R  :  /.'. 

Inscribe  in  the  ©  two  regular  polygons  of  the  same  number 
of  sides. 

Conceive  the  number  of  the  sides  of  these  similar  regular 
polygons  to  be  indefinitely  increased,  the  polygons  continuing  to 
be  inscribed,  and  to  have  the  same  number  of  sides. 

Then  the  perimeters  will  continue  to  have  the  same  ratio  as 
the  radii  of  their  circumscribed  circles,  §  374 

{the  perimeters  of  similar  regular  polygons  have  the  same  ratio  as  the  radii 
of  their  circumscribed  (D), 

and  will  approach  indefinitely  to  the  circumferences  as  their 
limits. 

.'.  the  circumferences  will  have  the  same  ratio  as  the  radii 
of  their  circles,  §  1 99 

.'.  C  :  C1  :  :  B  :  /,'. 

Q.  E.  D. 


REGULAR   POLYGONS    AND    CIRCLES.  217 

6.  Corollary.    By  multiplying  by  2,  both  terms  of  the 
ratio  R  \  R1,  we  have 

C  :  6"  :  :  2  R  :  2  R' ; 

that  is,  the  ei  mi  inferences  of  circles  are  to  each  other  as 
their  diami 


Since 


c 

:  C   : 

:  2/?  : 

2R', 

c 

:  27?  : 

:  C  : 

'2  R, 

2  R  s 

0 

'  2R' 

§262 


ratio  of  the  circumference  of  a  circle  to  its 
diameter  is  a  constant  quantity. 

tonstanl  quantity  is  denoted  by  the  Greek  letter**. 

377,  Scholium.  The  ratio  -  is  incommensurable,  and  there- 
in be  expressed    mly  approximately  in  Bgurea     The  let- 
In.  wever,  is  used  to  represent  its  exact  value. 


1.    Show  that   two  3  which  have  an  angle  of  the 

aqua]  to  the  supplement  of  the  angle  of  the  other  are  to  each 
other  as  the  products  of  the  sides  including  the  supplementary 
angli 

2.  Show,  geometrically,  that  the  square  described  upon  the 
sum  of  two  straight  lines  is  equivalent  to  the  sum  of  the  squares 

ribed  upon  the  two  lines  plus  twice  their  rectangle. 

3.  8h  >w,  geometrically,  that  the  square  described  upon  the 
difference  of  two  straight  lines  is  equivalent  to  the  sum  of  the 
squares  described  upon  the  two  lines  minus  twice  their  rectangle. 

L  Show,  geometrically,  that  the  rectangle  of  the  sum  and 
difference  of  two  Btraight  lines  is  equivalent  to  the  difference 
of  the  squares  on  these  lines. 


218  GEOMETRY. BOOK    V. 


Proposition  VIII.     Theorem. 

378.  If  the  number  of  sides  of  a  regular  inscribed  poly- 
gon be  increased  indefinitely ,  the  apothem  will  he  an  increas- 
ing variable  whose  limit  is  the  radius  of  the  circle. 


In  the  right  triangle  OCA,  let  0  A   be  denoted  by  R, 
OC  byr,  and  A  C  by  b. 

We  are  to  prove  Inn.  (r)  =  /?. 

r</,\  §52 

(a  J_  it  the  shortest  distance  from  a  point  to  a  straight  lim  ), 

An- 1  //-/•</,.  J97 

(one  side  of  a  A  is  greater  than  the  difference  of  the  other  two  tides). 

By  increasing  the  number  of  sides  of  the  polygon  indefi- 
nitely, A  1>,  thai  is,  '2  b,  can  be  made  Lobs  than  any  assigned 
quantity. 

.'.b,the  half  of  2  b,  can  be  made  less  than  any  assigned 
quantity. 

.'.R  —  r,  which  is  leu  than  b,  can  be  made  leas  than  any 

tied  quantity. 

.*.  /////.  (/,'  — /•)  =  <). 

.-.  R-  lim.  (r)«=0.  {  L99 

.*.  lim.  (r)  =  R. 

Q.  e.  d. 


REGULAR    POLYGONS    AND    CIRCLES. 


219 


Proposition  IX.     Theorem. 

379.   The  area  of  a  regular  polygon  is  equal  to  one-half 
thr  product  of  its  apothem  by  its  perimeter. 


Let  P  represent  the  perimeter  and  R    the  apothem 
of  the  regular  polygon  ABC,  etc. 

II  m      the  area  of  ABC,  etc.,  =  \  R  X  P. 

IMuu  OA%  OB,  OC,  etc. 

The  polygon  La  divided  Into  aa  many  A  as  it  has  sides. 

The  apothem  is  the  common  altitude  of  these  A, 

and    the   area    of  each   A   Lb  equal  to  \  R  multiplied  by 
the  h  §  324 

. ' .  i  ■  f  all  the  A  is  equal  to  \  R  multiplied  by  the 

.  of  all  the  baa 

But  the  sum  of  the  areas  of  all  the  A  is  equal  to  the  area 
of  the  poly 

and   the  sum   of  all  the  bases  of  the  A  is   equal  to  the 
perimeter  of  the  polygon. 

.*.  the  area  of  the  polygon  =  1-  R  X  P. 

Q.  E.  D. 


220  GEOMETRY.  BOOK    V. 


Proposition  X.     Theorem. 

380.    The   area    of  a   circle    is   equal  to   one-half  the 
product  of  its  radius  by  its  circumference. 


Let  R  represent  the  radius,  and  C  the  circumference 
of  a  circle. 

II V  are  to  prove      the  area  of  the  circle  =  |  R  X  G. 

[nscribe  any  regular  polygon,  and  denote  its  perimeter 
by  P,  and  its  apothem  by  r. 

Then  the  ana  erf  tbie  polygon  =  \  r  X  P,  §  379 

(the  area  of  a  regular  \  to  <>>t<  -Ik///  the  product  of  its  apothem 

by  th<  perimeter)* 

Conceive  the  number  of  Bides  of  this  polygon  to  be  indefi- 
nitely increased,  the  polygon  .-till  continuing  to  be  regular  and 
inscribed 

Then  the  perimeter  of  the  polygon  approaches  the  circum- 
ferenoe  of  the  circle  as  its  limit, 

the  apothem,  the  radius  as  its  limit,  §  37<S 

and  the  ana  of  the  polygon  approaches  the  O  as  its  limit. 

But  the  area  of  the  polygon  continues  to  be  equal  to  one- 
half  the  product  of  the  apothem  by  the  perimeter,  however 
great  the  number  of  sides  of  the  polygon. 

.-.  the  area  of  the  O  =  £  R  X  C.  §  1 W 

Q.  E.  D. 


REGULAR   POLYGONS    AND    CIRCLES. 


221 


381.  Corollary  1.    Since  =  tt, 

'2R 


§376 


.\C=2ttR. 
In  the  equality,  the  area  of  the  O  =  1 r  R  X  C, 
substitute  2  it  R  for  C  \ 
then  the  area  of  the  0  =  £#X27r7?, 
=  v  A'2. 
Thai  is,  the  area  of  a  O  =  it  times  the  square  on  its  radius. 

382.  Cor  2.     The  area  of  a  sector  equals  \  the  product  of 

'<  are;  for  the  sector  is  such  part  of  the  circle  as 
!•<•  is  of  tli»'  circumference. 

383.  I  > i ; i .    In  different  cir  .  similar  sectors, 
Mid  J                       ><fs,  are  such  as  correspond  to  equal  angles  at 

ntre, 

XI.    Theorem. 

384.  Two  circles  are  to  each   of  Iter  as  the  squares  on 

their  radii. 


Let  R  and  R'  be  the  radii  of  the  two  circles  Q  and  Q'. 
Q        R^ 

(?  =  tt7?2,  §381 

{the  area  of  a  O  =  w  times  (lie  square  on  its  radius), 

and  Q'  =  tt  R'2.  §  381 

Q         -  fii   __  R^ 

Q>  ~"  it  R'*  ~~  B* ' 

Q.  E.  D. 


—     = 


Then 


».  Corollary.    Similar  arcs,  being  like  parts  of  their  re- 
ive circumferences,  are  to  each  other  as  their  radii;  similar 
like  parts  of  their  respective  circles,  are  to  each 
/res  on  their  radii. 


2*22  GEOMETRY. BOOK    Y. 


Proposition  XII.     Theorem. 

386.  Similar  segments  are  to  each  other  as  the  squares 
on  their  radii.  C 

O  A 


P,  p 

Let  A  C  and  A'  C  be  the  radii  of  the  two  similar  seg- 
ments ABP  and  A' B' P'. 

w  .  ABP         AC1 

\\  e  are  to  prove       = . 

1  A'BhP'       jj~(jP- 

The  sectors  ACB  and  A'  C  B'  are  similar,  §  383 

(having  the  A  at  the  centre,  C  and  (7,  equal). 

In  the  A  ACB  and  A'C'B' 

/_C  =  ZC,  §  383 

(being  corresponding  A  of  similar  sectors). 

AC  =  CB,  §  1G3 

A'C'  =  C'B';  §  163 

.-.  the  A  A  C  11  and  A' C B'  are  similar,  §  284 

ing  an  Z  of  the  one  equal  to  an  Z  of  tlie  other,  and  the  including  sides 
proportional). 

Now  sector  ACB    _   AC2  §  385 

sector  A'C'B'        £Tj}? 
i  ilar  sectors  are  to  each  other  as  the  squares  on  their  radii) ; 

and  AACB   ,-£*  §342 

A  A'C'B'        JTq/*' 

(similar  &  are  to  each  other  as  the  squares  on  their  homologous  sides). 

jj  sector  ACB-  A  ACB   ■        ATO2 

Hence 


sector  A'  C  B'  -  A  A'  C  B'        ^HJ'2 
or, 


segment  A  B  P  J~C  &  271 

segment  A'  B'  P'        A7!!'2 
■■■■  quantities  be  increased  or  diminished  by  like  parts  of  each,  the  results 
will  be  in  Uie  same  ratio  as  the  quantities  themseh 

Q.  E.  D. 


EXERCISES.  223 


Exercises. 

1.  Show  that  an  equilateral  polygon  circumscribed  about  a 
circle  is  regular  if  the  number  of  its  sides  be  odd. 

2.  Show  that  an  equiangular  polygon  inscribed  in  a  circle  is 
ir  if  the  number  of  its  sides  be  odd. 

■  .    Show  that  any  equiangular  polygon  circumscribed  about  a 
circle  is  regular. 
vl    4.    Show  that  the  side  of  a  circumscribed  equilateral  triangle 
LB  double  the  side  of  an  inscribed  equilateral  triangle. 

5.    Show    that    the  area   of  a  regular  inscribed   hexagon  is 
ilis  of  that  of  the  regular  circumscribed  hexagon. 
Show  that  the  area  of  a  regular  inscribed  hexagon  is  a 
•  nil  1m!  areas  of  the  inscribed  and  cir- 

cumscribed equilateral  brian 

7.  Show  that  the  area  of  a  regular  inscribed  octagon  is  equal 
to  that  of  a  rectangle  whose  adjacent   sides   are  equal  to  the 

ibed  an' I  circumscribed  squares. 

8.  Show  that  the  area  <>t'  a  regular  inscribed  dodecagon  is 

te  square  on  the  radius. 

9.  Given  the  di  rff  a  circle  50;   find  the  area  of  the 

area  of  a  sector  of  80°  of  this  circle. 

1«>.  Three  equal  circles  touch  each  other  externally  and  thus 
inclose  one  acre  of  ground ;  find  the  radius  in  rods  of  each  of 
these  circles.  ' 

11.  Show  that  in  two  circles  of  different  radii,  angles  at  the 
centres  subtended  by  arcs  of  equal  length  are  to  each  other  in- 
versely as  the  radii. 

1 2.  Show  that  the  square  on  the  side  of  a  regular  inscribed 
pentagon,  minus  the  Bquare  on  the  side  of  a  regular  inscribed 
decagon,  is  equal  to  the  square  on  the  radius. 


224  GEOMETRY. BOOK    V. 


On  Constructions. 

Proposition  XIII.     Problem. 

387.    To  inscribe  a  tegular  jx>/t/gon  of  any  number  of 
tides  in  a  given  circle* 


"Let  Q  be  the  given  circle,  and  n  the  number  of  sides 
of  the  polygon. 

It  is  required  to  inscribe  in  Q,  a  regular  polygon  having  n 

Divide  the  circumference  of  the  O  into  n  equal  arcs. 

Join  the  extremities  of  these  .arcs. 

Then  we  have  the  polygon  required. 

For  the  polygon  is  equilateral,  §  181 

{in  the  same  O  equal  arcs  are  subtended  by  equal  chords)  ; 

and  the  polygon  is  also  regular,  §  364 

{an  equilateral  polygon  inscribed  in  a  O  is  regular). 

Q.  E.  F. 


CONSTRUCTIONS. 


225 


Proposition  XIV.     Problem. 

388.  To  inscribe  in  a  given  circle  a  regular  polygon 
which  has  double  the  number  of  sides  of  a  given  inscribed 
tegular  polygon. 


if 

Let  ABC  D  be  the  given  inscribed  polygon. 

It  is  required  to  in$cribe  a  regular  polygon  having  double  the 
N  u  mher  of  sides  of  A  BC  D. 

Bisect  the  arcs  A  B,  BC,  etc. 

v   A  /:.  KB,  BF,  etc., 

The  polygon  A  E  B  FC,  etc.,  is  the  polygon  required. 

the  chords  A  B,  BC,  etc.,  are  equal,  §  363 

(being  sides  of  polygon). 

.'.  the  arcs  A  B,  BC,  etc.,  are  equal,  §  182 

( M  the  same  O  equal  chords  subtend  equal  arcs). 

Hence  the  halves  of  these  arcs  are  equal, 

or,  AE,  E B,  BF,  FC,  etc.,  are  equal ; 

.'.  the  polygon  A  EB  F,  etc.,  is  equilateral. 

The  polygon  is  also  regular,  §  364 

(an  equilateral  polygon  inscribed  in  a  O  is  regular)  ; 

and  has  double  the  number  of  sides  of  the  given  regular 
polygon. 

Q.  E.  F. 


226 


GEOMETRY. BOOK    V. 


Proposition  XV.     Problem. 
389.  To  inscribe  a  square  in  a  given  circle. 


Let  0  be  the  centre  of  the  given  circle. 

It  is  required  to  inscribe  a  square  in  the  circle. 

Draw  the  two  diameters  A  C  and  B  D  JL  to  each  other. 

Join  AB,  BC,  CD,  and  DA. 

Then  A  B  C  D  is  the  square  required. 

For,  the  A  ABC,  BCD,  etc.,  are  rt.  A,  §  204 

(being  inscribed  in  a  semicircle) , 

and  the  sides  AB,  BC,  etc.,  ar^  equal,  §  181 

(in  the  same  O  equal  arcs  are  subtended  by  equal  chords)  ; 

.*.  the  figure  A  B  CD  is  a  square,  §  127 

(having  Us  sides  equal  and  its  A  rt.  A  ). 

Q.  E.  F. 

390.  Corollary.  By  bisecting  the  arcs  AB,  BC,  etc.,  a 
regular  polygon  of  8  sides  may  be  inscribed  ;  and,  by  continuing 
the  process,  regular  polygons  of  1G,  32,  G4,  etc.,  sides  may  be 
inscribed. 


>TRUCTIONS.  227 


Proposition  XVI.     Problem. 
891.   To  inscribe  in  a  given  circle  a  regular  hexagon 


v 


Let  0  be  the  centre  of  the  given  circle. 

It  is  required  to  inscribe  in  the  given  O  a  regular  hexagon. 

From  0  draw  any  radios,  as  OC. 

From  C  as  a  centre,  with  a  radius  equal  to  0  C, 

describe  an  arc  intersecting  the  circumference  at  F. 

Draw  OF*xi&  C  F. 

Then  C  F  \&  a  side  of  the  regular  hexagon  required. 

For  the  A  0  F  C  is  equilateral,  Cons. 

and  equiangular,  §  112 

.*.  the  Z  FO  C  is  J  of  2  rt.  A,  or,  £  of  4  rt.  A .        §  98 

.*.  the  arc  FC  is  \  of  the  circumference  A  B  C  F, 

.'.  the  chord  FC,  which  subtends  the  arc  FC,  is  a  side 
of  a  regular  hexagon  ; 

and  the  figure  CFD,  etc.,  formed  by  applying  the  radius 
six  times  as  a  chord,  is  the  hexagon  required. 

Q.  E.  F. 

392.  Corollary  1.  By  joining  the  alternate  vertices  A,  C, 
/),  an  equilateral  A  is  inscribed  in  a  circle. 

393.  Cor.  2.  By  bisecting  the  arcs  A  B,  B  C,  etc.,  a  regu- 
lar polygon  of  12  sides  may  be  inscribed  in  a  circle ;  and,  by 
continuing  the  process,  regular  polygons  of  24,  48,  etc.,  sides 
may  be  inscribed. 


228 


GEOMETRY.  —  BOOK    V. 


Proposition  XVII.     Problem. 
394.   To  inscribe  in  a  given  circle  a  regular  decagon. 


Let  0  be  the  centre  of  the  given  circle. 
If  is  required  tot  tki  given  O  a  regular  decagon. 

Draw  the  radius  0  0, 

and  divide  it  in  extreme  and  mean  ratio,  so  that  0  0  shall 
be  to  OS  as  0  S  is  to  SO.  §  311 

From  0  as  a  centre,  -with  a  radius  equal  to  OS, 
describe  an  arc  intersecting  the  circumference  at  B. 

Draw  BO,  BS,  and  B 0. 
Then  B  0  is  a  side  of  the  regular  decagon  required. 


For 

and 


00 


( Jons. 
Cons. 


then 


OS  :  :  OS  :  SO, 

BO=0  8. 

Substitute  for  0  S  its  equal  B  0, 

0  0  :  BO  ::  BO  :  SO. 

Moreover  the  Z  0  0  B  =  Z  S  0  B, 

.'.the  A  00  B  and  B  OS  are  similar, 

ng  an  A  oftheone  equal  to  an  Z.  of  the  other,  and  the  including  sides 
proportional). 

But  the  A  00 B  is  i  §  160 

(ifs  sides  0  C  and  OB  being  radii  of  the  same  circle). 

.'.  the  ABO  S,  which  is  similar  to  the  A  0  OB,  is  isoscel 


Id  en. 
§284 


CONSTRUCTIONS.  229 

and                              BS  =  BG.  §114 

Bat                               OS  =  BC,  Cons. 

.'.OS  =  BS,  Ax.  1 

■\  the  A  S  0  B  is  isosceles, 

and  theZ  0  =  Z  S  B  0,  §112 

(tein#  opposite  equal  sides). 

But  the  ZCSB  =  ZO+ZSBO,  §  105 

estcrior  Z  of  a  A  is  t  nf  the  two  opposite  interior  A  ). 

.\theZ  CSB  =  2Z  0. 
Z  SO B(=  Z  C8B)  °~2Z  0,  §  112 

and  Z  OBC  (=  Z  SCJi)  =  2  Z  0.  §112 

.-.  the  sum  of  the  <d  of  the  A  OCB  =  5  Z  0. 

/.  5  Z  0  =  2rt.  ^,  §98 

and         Z  0  =  £  of  2  rt.  zi,  or  ^  of  4  rt.  4. 

.'.  the  arc  2?  (7  is  -^  of  the  circumference,  and 
.-.  the  chord  BC  is  a  side  of  a  regular  inscribed  decagon. 

Bence,  to  inscribe  a  regular  decagon,  divide  the  radius  in 
and    mean  ratio,   and   apply  the  greater  segment  ten 
times  as  a  chord. 

Q.  E.  F. 

395.  Corollary  1.    By  joining  the  alternate  vertices  of  a 
liar  inscribed  decagon,  a  regular  pentagon  may  be  inscribed. 

39G.  Cor.  2.  By  bisecting  the  arcs  BC,  C F,  etc.,  a  regular 
polygon  of  20  sides  may  be  inscribed,  and,  by  continuing  the 
process,  regular  polygons  of  40,  80,  etc.,  sides  may  be  inscribed. 


:i30 


GEOMETRY. BOOK    V. 


Proposition  XYIII.     Problem. 

397.   To  inscribe  in  a  given  circle  a  regular  pentedecagon, 
or  polygon  of  fifteen  sides. 


Let  Q  be  the  given  circle. 
It  is  required  to  inscribe  in  Q  a  regular  \  ton. 

Draw  EH  equal  I  I  a  regular  inscribed  hexagon,    §  391 

and  EF  equal  to  a  aid  igular  inscribed  decagon.     §  39  1 

Join  FH. 
Then  FH  will  be  a  side  of  a  regular  inscribed  pentedecagon. 
For        the  arc  EH  La  |  of  the  circumference, 

and  the  arc  A'/*' is  {\t  of  the  circumference; 
.*.  the  are  /•'//  is  \  —  ^  or  ^,  of  the  circumference. 
.*.  the  chord  /•'//  is  a  aide  of  a  regular  Inscribed  pente- 

and  by  applying  FH  fifteen  times  as  a  chord,  we  have  the 
polygon  requii 

Q.  E.  F. 

398.  Corollary.  By  bisecting  the  arcs  FIf,  If  A,  etc., 
a  regular  polygon  of  30  sides  may  be  inscribed  ;  and  by  con- 
tinuing the  pr  dar  polygons  of  GO,  120,  etc.  sides  may 
be  inscribed. 


BT&UCTIOl  231 


Proposition  XIX.     Problem. 

399.   To   inscribe  in  a  given  circle  a  regular  polygon 
nmilar  (<>  a  give*  regular  polygon. 

J9'  CD 


Let  A  BCD,   etc.,   be  the  given  regular  polygon,  and 
C  D'  E'  the  given  circle. 

ft    m   required  tc  C  D'  E'   a  regular  polygov 

si  in  i  In  r  t<>  A  B  C  D. 

mi  (a  the  centre  <-i*  the  polygon  A  BC D,  etc. 

draw  OB  and  OC. 

•in  0'  the  centra  of  the  O  C D'  &t 

draw  VCsad  o1  D\ 

making  the  /.<)'  =  £  0. 

Draw  6"  7/. 

Then  CD'  will  be  a  side  of  the  regular  polygon  required. 

For   each    polygon  will   have  as  many  sides  as  the  A  0 
(=Z  0')  is  contained  times  in  4  rt.  A. 

.'.    the    polygon    <u  D' E',   etc.    is    similar  to  the  polygon 
r/y /;.  etc.,  §  372 

{two  regular  polygons  o/tkt  same  number  of  sides  are  similar). 

Q.  E.  F. 


'Sd'2  GEOMETRY. BOOK    V. 


Proposition  XX.     Problem. 

400.   To  circumscribe  about  a  circle  a  regular  polygon 
similar  to  a  given  inscribed  regular  polygon. 


BMC 

Ha 

/?T%# 

y 

"""-.    /V 

pV 

.....*:4'.'. aL<? 

A 

/  \  k 

N 
Let  II MRS,  etc.,  be  a  given  inscribed  regular  polygon. 

It    is  required  to  circumscribe  a  regular  polygon   similar 
to  H  M  RS,  etc. 

At  the  vertices  H,  M,  R,  etc.,  draw  tangents  to  the  O, 
intersecting  each  other  at  A,  B,  C,  etc. 

Then  the  polygon  ABC  D,  etc.  will  be  the  regular  poly- 
gon required. 

Since  the  polygon  ABC D,  etc. 

has  the  same  number  of  sides  as  the  polygon  If  MRS,  etc., 

it  is  only  necessary  to  prove  that  ABC D,  etc.  is  a  regular 
polygon.  §  372 

In  the  A  B H M  and  CMS, 

II  M  -  M  /,'.  §  363 

(being  sulcs  of  a  regular  polygon), 


I  INSTRUCTIONS.  233 


the  A  B II JA  B  M  IL  C  M  /?,  and  C  R  M  are  equal,     §  209 
ng  measured  by  luilves  of  equal  arcs)  ; 

.\  the  A  BUM  and  CMR  are  equal,  §  107 

i  adja&  at  A  <>/  the  one  equal  respectively  to  a  side  and 
cent  A  of  the  otJier). 

(being  homologous  A  of  equal  &  ). 

In  like  maimer  ire  may  prove  Z  C  =  Z.  D,  etc 

.-.  the  polygon  .1  BCD,  etc.,  is  equiangular. 

Since  the  ...  B  II  M,  CM  R,  etc,  are  isosceles,       §  241 

///  to  a  Q.arc  equal), 

the  sidee  A'//,  A'JA  Clf,  CRt  etc,  are  equal, 

{being  homologous  sides  of  equal  isosceles  &). 

.  .  the  sides  AB>BC,CD,  etc.  are  equal,         Ax.  6 

and  the  polygon  A  BC  Dt  etc  La  equilateral 

Therefore  the  circumscribed  polygon  is  regular  and  similar 
to  the  given  inscribed  polygon.  §  372 

Q.  E  F. 


A*  denote  the  radius  of  a  regular  inscribed  polygon, 
/•  the  apothem,  a  one  side,  A  one  angle,  and  C  the  angle  at  the 

(■••litre  ;   Bh0W  that 

1.  In    a    tegular    inscribed    triangle    a  =  R  V^,    r  =  \  R, 
A  =  60°,  C  =  120°. 

2.  In  an  inscribed  equate  a  =  R  v7^,  r  =*  \R  ^2,  A  =  90°, 
C  =  90°. 

3.  In   a   regular   inscribed    hexagon   a  =  R,    r  =  ^  R  V^3, 
A  =  120°,  (7  =  60°. 

R  0/5  -  1) 

I.    In    a    regular    inscribed    decagon     a   =  g > 

r  =  J  /*  ^10  +  2  V6,   ^  =  U4°,    (7  =  36°. 


234  GEOMETRY. BOOK    V. 


Proposition  XXI.     Problem. 

401.   To  find  the  value  of  the  chord  of  one-half  an  arc, 

in  terms  of  the  chord  of  the  whole  arc  and  the  radius  oj  the 

circle, 

D 


Let  A  B  be  the  chord  of  arc  A  B  and  A  D  the  chord 
of  one- half  the  arc  A  />. 

It  is  required  to  find  the  value  of  A  D  in  terms  of  A  B  and 
11  (radius). 

From  D  draw  J)  II  through  the  centre  0, 

and  draw  0  A. 

II  I)  is  J_  to  the  choid  A  B  at  its  middle  point  C,      §  00 

{two  points,  0  and  D,  equally  distant  from  the  extremities,  A  and  B,  de- 
termine the  position  of  a  ±  to  the  middle  point  of  A  B). 

The  Z  HA  D  is  a  rt.  Z,  §  204 

/  inscribed  in  a  semicircle), 

.\A~Jf  =  DHX  DC,  §289 

{tlie  square  on  one  side  of  a  rt.  A  is  equal  to  /In'  product  of  the  hypotenuse  by 
the  adjacent  segment  made  by  the  J_  let  fall  from  the  vertex  of  tJie  rt.  Z  ). 

Now  D  II  =  2  //, 

and  DC  =  DO-CO  =  B-CO; 

.\ADl  -  -111(11-  CO). 


■  0N8TBUCTION8.  235 


Since  A  C  0  is  a  rt.  A, 

i  <r  =  A~C?  +  CO2 ;  §331 

.-.07/  =  Xtf-Ai?. 


C0  =  ^(A~Oi-A(f), 


->!'■ 


4  A-  -  AB\ 

=  V4  IP  -  AR*. 
2 

h.  the  equation  AH?  =  2R  (R—  CO), 


substitute 

for  C  0  its  value  V4  *»  -  ^ 
2 

then 

V                     2 
2  8»  — W^4JP- 

-J-5'V 

.  J  0  -  \    2  A'-  -  R  NilP-  A  tf\  . 

Q.  E.  F. 

(  loROLLART.     It*  \\v  take  the  radius  equal  to  unity, 


equation  A  D  -  t/-J  A'-  -  £  ^4  #2  -  i ~i*2)  becomes 
AD  =  ^2-  V^-Jtf2. 


236 


GEOMETRY. BOOK    V. 


Proposition  XXII.     Problem. 

403.   To   compute   the   ratio   of  the  circumference  of  a 
circle  to  its  diameter,  approximately. 


Since 


§370 


Let   C   be   the   circumference  and   R    the  radius  of  a 
circle. 

,=    G 
2  It' 

C 

Wll.'ll    R  =    1,    7T  =   -  ' 

//  i$  required  to  find  the  numerical  ''due  of  tt. 

We  make   the  following  -computations  by  the  use  <>1  tqe 


l 


formula  obtained  in  the  last  proposition, 


Nh. 


A  1)  =  1/2  -  V^4  -  ii  2?2, 

when  .4  /?  is  a  side  of  a  regular  hexagon  : 
In  a  polygon  of 

Form  of  Computation.                   Length  of  Bide.  Perimeter. 

12     .1 1)  =  y/2  -  y/J^P .517G3809  G.211G5708 

2  I     J  />  =  ^2-^4  — (.51763809)2     .26105238  6.26525722 

H     .-1  /)  =  ^2^4  -(.26 105238^     .13080626  6.27870041 

96     A  I)  —  ^2  —  ^4  — (.13080626)2     .06543817  6.28206396 

192     AD  =  ^2  —  ^i  —  (.06543*  I  7  g     .08272346  6.28290510 

AD  =  >]'2  —  )Ji  —  (.0327234^     .01636228  6.28311544 

768     A  D  =  ^2  -  V4  —  (.01636228)2     .00818121  6.28816941 

Eenoe  we  may  consider  6.28317  as  approximately  the  cir 
cumference  of  a  O  whose  radius  is  unity. 


7r,  which  equals  —  , 


6.28317 

2 


.-.  7t  =  3.1  U59  nearly. 


ISOPERIMETRICAL    POLYGONS.  237 


On  Isoperimetrical  Polygons.  —  Supplementary. 

H' L    I)i;f.    Isoperimetrical  figures  are  figures  which  have 
aqua]  perimet 

>.  Def.    Among   magnitudes   of    the    same    kind,    that 
whirl i   is  greatest  is  a  Maximum,   and  that  which  is  smallest 

Minimum, 

I'h us  the  diameter  of  a  circle  is  the  maximum  among  all 
inscribed  straight  lines;  and  a  perpendicular  is  the  minimum 

among  all  straight  lines  drawn  from  a  point  to  a  given  straight 
line. 

Proposition  XXIII.     Theori 

406.  Of  all  triangle*  having  two  ride*  respectively  equal, 

f/taf   in    wkiek   these  B  .il'f  angle  is  the  maxi- 

A 
E  hv       E 


Let  the  triangles  A  BC  and  EBC  have  the  sides  AB 
and  BC  equal  respectively  to  KB  and  BC]  and 
let  the  angle  ABC  be  a  right  angle. 

m      AABOAEBC. 
From  A'.  Lei  fall  the  _L  E  D. 

The  A  ABC  and  EBC,  having  the  same  base  B  C,  are  to 
each  other  as  their  altitudes  A  B  and  ED,  §  326 

(&  }tn  /se  are  to  each  otJier  as  their  altitudes). 

Nnw  ED  is  <EB,  §52 

(a  _L  is  the  shortest  distance  from  a-point  to  a  straight  line). 
But  EB  =  AB,  Hyp. 

.-.  ED\*<AB. 
.'.A  ABO  A  EBC. 

Q.  E.  D. 


238  GEOMETRY. 


Proposition  XXIY.     Theorem. 

407.  Of  all  polygons  formed  of  sides  all  given,  but  one, 
the  polygon  inscribed  in  a  semicircle,  having  the  undetermined 
side  for  its  diameter,  is  the  maximum. 


Let  A  B,  BC,  CD,  and  I)  E  be  the  sides  of  a  polygon 
inscribed  in  a  semicircle  having  A  E  for  its  di- 
ameter. 

II  <    are  to  prove  the  polygon  ABODE  the  maximum  of 
fans  homing  the  sides  A  B,  BO,  CD,  and  D E. 

From  any  vertex,  as  0,  draw  C  A  and  C  E. 

Then  the  Z  A  CE  is  a  rt.  Z  ,  §  204 

(being  inscribed  in  a  semicircle). 

Now  the  polygon  is  divided  into  three  parts,  ABC,  CD  E, 
and  A  C  K. 

Tin-  parts  ABC  and  CDE  will  remain  the  same,  if  the 
Z  A  C  E  be  increased  or  diminished  ; 

but  the  part  ACE  will  be  diminished,  §  406 

(of  all  &  having  two  sides  respectively  equal,  that  in  which  these  sides  in- 
clude art.  Z  is  the  maximum). 

.'.  A  B  C D  E  is  the  maximum  polygon. 

Q.  E.  D. 


ISOPERIMETRICAL    POLYGONS. 


239 


Proposition  XXV.     Theorem. 

408.  The  maanmum  <f  all  polygons  formed  of  given  sides 
in  be  inscribed  in  a  circle, 

W      D> 
A 


A  A' 

Let  A  BCDB  be  a  polygon  inscribed  in  a  circle,  and 
A'  I> '  <"  J)'  /;•'  be  a  polygon,  equilateral  with  re- 
spect to  ABODE,  but  which  cannot  be  inscribed 
in  a  circle. 

We  are  to  prove 
the  polygon  A  It  C  D  E  >  the  polygon  A' B' Cl  D'  E'. 
Draw  the  diameter  A  II. 
Join  C// and  J)  II. 
:i  C D1  (=CD)  construct  the  A  C II'  D'  =  A  C  H Dy 
and  draw  A1  II'. 
Xuw  the  polygon  A  BOB  >  the  polygon  A'B'C'IP,     §  407 

fans  fonncd  of  sides  all  given  but  one,  the  polygon  inscribed  in  a 
tied  side,  for  its  diameter,  is  the  maximum). 

And  the  polygon  A  E  D  II  >  the  polygon  A'  E'  D'  H'.     §  407 

Add  these  two  inequalities,  then 

the  polygon  ABC II BE  >  the  polygon  A' Bf C Hl D1  E' . 

Take  away   from  the  two  figures  the  equal  A  CHD  and 

t  ■  II  //. 

Then  the  polygon  ABCDE>the  polygon  A* B' C D' E'. 

Q.  E.  D 


240 


GEOMETRY. BOOK  V. 


Proposition  XXVI.     Theorem. 

409.    Of  all  triangles  hiving  the  same  base  and  equal 
perimeters,  the  isosceles  triangle  is  the  maximum. 


,\II 


Let    the   AACB   and  ADB  have    equal  perimeters, 
and  let  the  AACB  be  isosceles. 

FPi  are  to  prove      AACB>AADB. 

Draw  the  J§  OJBmd  1>  I'. 

AACB        CE 

AABD        1)F  y 

(&  having  the  same  bast  err  fa  ttuk  other  as  their  altitudes). 

Produce  A  C  to  //,  making  Off**  AC. 

D»W  1/  II. 

The  Z  A  B  JI  is  a  rt.  Z,  for  it  will   be  inscribed   in  the 
semicircle  drawn  from  C  as  a  centre,  with  fch<  radios  CB. 


ISOPERIMETRICAL    POLYGONS.  241 


From  (7  let  fall  the  LCK; 
and  from  D  as  a  centre,  with  a  radius  equal  to  D  B, 
describe  an  arc  cutting  //  B  produced,  at  ^. 
Draw  D  P  and  A  P, 
and  let  fall  the ±  DM. 
Since         .\!l  =  AC+CB  =  AD  +  DB, 
and  AP<AI)+  DP-, 

.'.  A  P<AD  +  DB; 
.-.  A  //>  A  P. 
.-.  BH>  BP.  §56 

BK  =  \BH,  §113 

(a  _L  tfnriOTi  /rowi  /Ac  rcrtec  o/a»  isosceles  A  fcisccte  /Ac  tec)> 

and  BM=hBP.  §113 

But  '•/-'     »*;  §185 

(II*  compi-chcndcd  between  \\s  are  equal); 
and  DF=BM,  §136 

.-.  CE>  DF. 
.:AACB>  AADB. 

Q.  E.  D. 


242 


GEOMETRY. BOOK    V. 


Proposition  XXVII.     Theorem. 

410.   The  maximum  of  isoperimetrical  polygons  of  the 
same  number  of  sides  is  equilateral. 


Let  ABCD,  etc.,  be  the  maximum  of  isoperimetrical 
polygons  of  any  given  number  of  sides. 

If,  an  to  prove      A  B,  BC,  C  D,  etc.,  equal. 

Draw  A  C. 

The  A  ABC  must  be  the  maximum  of  all  the  A  which 
are  formed  upon  A  C  with  a  perimeter  equal  to  that  of  A  ABC. 

Otherwise,  a  greater  A  A  KC  could  be  substituted  for  A  ABC, 
without  changing  the  perimeter  of  the  polygon. 

But  this  is  inconsistent  with  the  hypothesis  that  the  poly- 
gon ABCD,  etc.,  is  the  maximum  polygon. 

.*.  the  A  A  B  C,  is  isosceles,  §  409 

{<>/  all  A  having  the  same  base  and  equal  perimeters,  tlie  isosceles  A  is  the 
maximum  ». 

In  like  manner  it  may  be  proved  that  B  C =  CD,  etc. 

Q.  E.  D. 

111.  Corollary.  The  maximum  of  isoperimetrical  poly- 
gons of  the  same  number  of  sides  is  a  regular  polygon. 

For,  it  is  equilateral,  §  410 

(the  maximum  of  isoperimetrical  polygons  of  the  same  number  of  tides  is 
equilateral). 

Also  it  can  be  inscribed  in  a  O,  §  408 

(the  maximum  of  all  polygons  formed  of  given  sides  can  be  inscribed  in  a  O). 

Hence  it  is  regular,  §  364 

(tin  equilateral  polygon  inscribed  fa  a  0  is  regular). 


ISOPERIMETRICAL    POLYGONS. 


243 


Proposition  XXYIII.     Theorem. 

412.    Ofiiqperimeirieal  regular  polygons,  that  is  greatest 
7/  ias  the  greatest  number  of  ride*. 


Let  Q  be  a  regular  polygon  of  three  sides,  and  Q'  be 
a  regular  polygon  of  four  sides,  each  having  the 
same  perimeter. 

We  are  to  prove      Q'  >  Q. 

1  n  any  side  A  B  of  Q,  take  any  point  D. 

The   polj  may  be  considered  an   irregular  polygon 

of  fi  in  which  the  sides  A  D  and  D  B  make  with  each 

r  an  A  equal  to  two  rt.  A . 

Then  the  irregular  polygon  Q,  of  four  sides  is  less  than  the 

r  isoperimetrical  polygon  Q'  of  four  sides,  §  411 

naximwm  of  isoperimetrical  polygons  of  the  same  number  of  sides  is  a 
regular  polygon). 

In  like  manner  it  may  be  shown   that  Qf  is  less  than  a 
regular  isoperimetrical  polygon  of  five  sides,  and  so  on. 

Q.  E.  D. 


413.  Corollary.    Of  all  isoperimetrical  plane  figures  the 
circle  ifl  the  maximum. 


244 


GEOMETRY. 


BOOK    Y. 


Proposition  XXIX.     Theorem. 

414.  If  a  regular  polygon  be  constructed  with  a  given 
area,  its  perimeter  will  be  the  less  the  greater  the  number 
of  its  sides. 


Let  Q  and  Q'  be  regular  polygons   having   the   same 
area,  and  let  Q'  have  the  greater  number  of  sides. 

We  are  to  prove  the  perimeter  of  Q>  the  perimeter  of  Q'. 

Let  Q"  be  a  regular  polygon  having  the  same  perimeter  as 
Q'y  and  the  same  number  of  sides  as  Q. 

Th.-n  Cis><?",  §412 

(of  isopcrimctrical  regular  ])olyg<>  us,  thai  is  t In greatest  which  has  the  greatest 

>/>'S). 

But  Q  =  Q', 

.'.  £is><2". 
.*.  the  perimeter  of  Q  is  >  the  perimeter  of  Q". 
But  the  perimeter  of  Q'  =  the  perimeter  of  Q",      Cons. 
.-.  the  perimeter  of  Q  is  >  that  of  Q'. 

Q.  E.  D. 

415.  Corolla  by.    The  circumference  of  a  circle  is  less  than 
the  perimeter  of  any  other  plane  figure  of  equal  area. 


SYMMETRY. 


245 


On  Symmetry.  —  Supplementary. 

416.  Two  points  are  Symmetrical  when  they  are  situated 
nil  opposite  sides  of,  and  at  equal  distance* ./font,  a  fixed  point, 
line,  or  plane,  taken  as  an  object  of  reference. 

417.  When  a  point  is  taken  as  an  object  of  reference,  it  is 
called  the  Centre  of  Symmetry  .  when  a  line  is  taken,  it  is  called 
the  Axis  of  Symmetry ;  when  a  plane  is  taken,  it  is  called  the 
Pin  ry. 

418.  Two  points  are  symmetrical  with  re- 
spect to  b  autre  bisect  the  sti. 
tine  terminated  by  these  points.     Thus,  J\  p 
are  symmetrica]  with  respect  to  C,  if  C  b 
the  straight  line  PP. 

419.  The  distance  of  either  of  the  two  symmetrical  points 
from  the  called  the  Rodin  s  of  Symmetry, 

Thus  either  OP  01  CP  ifl  the  radius  of  symmetry. 


420.     Txoo   points   are    symmetrieal    with 
set  to  an  axis,  if  the  axis  bisect  at  right 
-    the    straight    line    terminated    by  these 
points.     Thus,  1\  P  are  symmetrical  with  re- 
in the  axis  XX',  if  XX'  bisect  PP  at 
right  angles. 


1 2 1 .  Two  points  are  symmetrical  with 
ct  to  a  plane,  if  the  plane  bisect  at 
right  angles  the  straight  line  terminated  by 
these  points.  Tims  P,  P'  are  symmetrical 
with  respect  to  MN,  if  M  N  bisect  P  P'  at 
right  angles. 


ML 


Pt 


rN 


246 


GEOMETRY. 


-BOOK    V. 


422.  Two  plane  figures  are  symmetrical  with  respect  to  a 
centre,  an  axis,  or  a  plane,  if  every  point  of  either  figure  have 
its  corresponding  symmetrical  point  in  the  other. 

B> 


Fig.  1. 


Fig.  2. 


Fig.  3. 


Thus,  the  lines  A  B  and  A'  B'  are  symmetrical  with  respect 
to  the  centre  C  (Fig.  1),  to  the  axis  X  X'  (Fig.  2),  to  the  plane 
MX  (Fig.  3),  if  every  point  of  either  have  its  corresponding 
symmetrical  point  in  the  other. 


V        I 


c< 


.V 


W     D' 


ML 


Fig.  6. 

Also,  the  triangles  A  B  D  and  A'  B1  D'  are  symmetrical  with 

■t  to  the  centre  C  (Fig.  4),  to  the  axis  XX'  (Fig.  5),  to  the 

plane  MX  (Fig.  G),  if  every  point  in  the  perimeter  of  either 

have  it.s  corresponding  symmetrica]  point  in  the  perimeter  of  the 

other. 

423.   Def.    In  two  symmetrical  figures  the  corresponding 
Symmetrical  points  and  lines  arc  called  homologous. 


SYMMETRY. 


247 


Two  symmetrical  figures  with  respect  to  a  centre  can  be 
brought  into  coincidence  by  revolving  one  of  them  in  its  own 
plane  about  the  centre,  every  radius  of  symmetry  revolving 
through  two  right  angles  at  the  same  time. 

Two  symmetrica]  figures  with  respect  to  an  axis  can  be 
brought  into  coincidence  by  the  revolution  of  either  about  the 
axis  until  it  conns  into  the  plane  of  the  other. 

424.  Dip.  A  single  figure  is  a  lymmetrical  figure,  either 
when  it  can  be  divided  by  an  axis,  or  plane,  into  two  figures 
symmetrica]  with   respect    to  that  axis  or  plane;  or,  when  it  has 

atre  such  that  every  straight  line  drawn  through  it  cuts  the 
perimeter  of  the  figure  in  two  points  which  are  symmetrical 
with  respect  to  that   centre. 


Fig.  1. 


Thus,  Fig.  1  is  a  symmetrica]  figure  with  respect  to  the 
divided  by  X  A"  into  figures  ABC D  and  AB'C'D 
which  are  symmetrica]  with  reaped  to  XX'. 

And,   Fi-  2  is  a  symmetrica]  figure  with  respect  to  the 

e  0,  if  the  centre  0  bisect  every  straight  line  drawn 
through  it  and  terminated  by  the  perimeter. 

v  such  straight  line  is  called  a  diameter. 

The  circle  is  an  illustration  of  a  single  figure  symmetrical 
with  respect  to  its  centre  as  the  centre  of  symmetry,  or  to  any 
diai;.  of  symmetry. 


248  GEOMETRY. BOOK    V. 


Proposition  XXX.     Theorem. 

1:25.   Two  equal  and  parallel  lines  are  symmetrical  with 
respect  to  a  centre. 

A  B> 


B  A' 

Let  AB  and  A1  B'  be  equal  and  parallel  lines. 

We  are  to  prove      A  B  and  A'  B'  symmetrical. 

Draw  A  A'  and  B  />',  and  through  the  point  of  their  inter- 
section C,  draw  any  other  line  11(11',  terminated  in  AB  and 
A'B'. 

In  the  A  CAB  md  C  A'  h> 

A  11  =  A'B',  Hyp. 

also,      A  A  and  B  =  A  A'  and  B'  respectively,  §68 

(firing  off. -inf.  A), 

.-.A  CAB  =  A  CA'B';  §  107 

,\  CA  and  CB  =  CA'  and  C  B'  respectively, 
(//>  ///'/  homologoxu  sides  of  equal  &). 

Now  in  the  A  A  C II  and  A' C II' 

AC  =  A'C, 

A  A  and  A  C  H '■■  A  A'  and  A'C  H*  respectively, 

.'.A  A  GII  =  A  A'C  IV,  §  107 

[hewing  a  ride  and  two  adjj,  A  of  the  one  equal  respectively  to  a  side  and  two 
adj.  A  of  the  other), 

.'.  CII=G  //', 

(being  homologous  sides  of  equal  A ). 

.*.  //'  is  the  symmetrical  point  of  //. 

r»»it  //  is  any  point  in  A  B ; 

.'.  every  point  in  A  B  has  its  symmetrica]  point  in  A' B'. 

.*.  A  B  and  A' B'  are  symmetrical  with  respect  to  G  as  a 
centre  of  symmetry. 

Q.  E.  D. 

426.  Corollary.  If  the  extremities  of  one  line  be  re- 
spectively the  symmetricals  of  another  line  with  respect  to  the 
same  centre,  the  two  lines  are  symmetrical  with   respect  to  that 

centre. 


SYMMETRY. 


249 


Proposition  XXXI.     Theorem. 

I*7«   Tf  a  figure  he  q  //  with  respect  to  two  axes 

vendicular  to  each  other,  it  is  symmetrical  with  respect 
to  tin  w  as  a  centre. 

Y 


JD 


V 

Let  the  figure  ABODE FO //   be  symmetrical    to    the 

two  axes  X  X'%    )    V  which  intersect  at  0. 

Wi  0  u  try  of  the  figure, 

I  be  any  pouri  in  the  perimeter  of  the  figure. 

Draw  IKL  ±  I  tid  IMN±  to  )'  V. 

J«n  £0,  o.\\  and  AM/. 

Now  a/-  /■. 

7  /////i  respect  to  X  X!). 

Bui  A'/-    02f, 

(lis  comprehended  between  \\s  ar 
.\  A' A  -  0JT. 
.-.  KLOM  k  a  O, 

(having  two  sides  equal  and  parulhl). 
.\  LO  is  equal  and  parallel  to  Ar  J/, 
'<7  opposite  sides  of  a  CJ). 
In  like  manner  we  may  prove  OiV  equal  and  parallel  to  AT  if. 
Hence  the  points  Z,  0,  and  ^V  are  in  the  same  straight  line 
drawn  through  the  point  0  IJ  to  KM. 
Also  L0  =  0N, 

{since  each  is  equal  to  KM). 
lit  line  L  0  X,  drawn  through  0,  is  bisected  at  0. 
.'.  0  is  the  centre  of  symmetry  of  the  figure.        §  424 

O.  E.  D. 


§420 

§  135 

Ax.  1 
§  136 

§  134 


250 


GEOMETRY. 


BOOK    V. 


Exercises. 

1.  The  area  of  any  triangle  may  be  found  as  follows  :  From 
half  the  sum  of  the  three  sides  subtract  each  side  severally,  mul- 
tiply together  the  half  sum  and  the  three  remainders,  and  extract 
the  square  root  of  the  product. 

Denote  the  sides  of  the  tri- 
angle A  B  C  by  a,  b,  c,  the  alti- 

ii                       f'l  +  b  +  c  . 
tude  by  p,  and by  s. 

Show  that 

a2  =  b2+-c*-'2cXAD, 

AD 

and  show  that 

i  * 


P 


2c 


p  = 


=  V  (b+c  +  a)(b  +  c—  a)  (a  +  b~c)  (a—  b  +  c) 


2e 


Hence,  show  that  area  of  A  A  B  0,  which  La  equal  to 


t  X  p 


=  \<J  {b  +  c  +  a)  (b  +  c-  a)  {a  +  b-c)  (a-b  +  c), 
=  ^  s(s  —  a)(s  —  b)(s  —  c). 

2.  Show  that  the  area  of  an  equilateral  triangle,  each  side  of 
which  is  denoted  by  a,  is  equal  to  — j—  . 

3.  How  many  acres  are  contained  in  a  triangle  whose  sides 

lively  GO,  70,  and  80  chains? 

4.  How  many  feet  are  contained  in  a  triangle   each  side  Oi 
which  La  75  feet? 


G 


THIS  BOOK  IS  DUE  ON  THE  LAST  DATE 
STAMPED  BELOW 


AN  INITIAL  FINE  OF  25  CENTS 

WILL  BE  ASSESSED  FOR  FAILURE  TO  RETURN 
THIS  BOOK  ON  THE  DATE  DUE.  THE  PENALTY 
WILL  INCREASE  TO  SO  CENTS  ON  THE  FOURTH 
DAY  AND  TO  $1.00  ON  THE  SEVENTH  DAY 
OVERDUE. 


MftY  13  v 


CT  14  1937 


fEB    7  \m 


^3**- 


20Mar'fifiEQJ 


FEB    4  1939 


«W  Cf 


NOV   6 


iflai 


*7 


OCT  27   1942 


RECD  LD 


1  J-  U  ?•'  "J|V" 





MAR    10  1943 


MAY  4-  1966  3  0 


to26lft<»0RCD 


»   1C  1Q44 


LD  21-95m-7,'37 


VB  66620 


